Business & Data Analytics Discussion

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GAME AT MCDONALD’S Several years ago, McDonald’s ran a campaign in which it gave game cards to its customers. These game cards made it possible for customers to win hamburgers, french fries, soft drinks, and other fast-food items, as well as cash prizes. Each card had 10 covered spots that could be uncovered by rubbing them with a coin. Beneath three of these spots were “zaps.” Beneath the other seven spots were names of prizes, two of which were identical. For example, one card might have two pictures of a hamburger, one picture of a Coke, one of french fries, one of a milk shake, one of $5, one of $1000, and three zaps. For this card the customer could win a ham- burger. To win on any card, the customer had to uncover the two matching spots (which showed the potential prize for that

card) before uncovering a zap; any card with a zap uncovered was automatically void. Assuming that the two matches and the three zaps were arranged randomly on the cards, what is the probability of a customer winning?

We label the two matching spots M1 and M2, and the three zaps Z1, Z2, and Z3. Then the probability of winning is the probability of uncovering M1 and M2 before uncovering Z1, Z2, or Z3. In this case the relevant set of outcomes is the set of all orderings of M1, M2, Z1, Z2, and Z3, shown in the order they are uncovered. As far as the outcome of the game is concerned, the other five spots on the card are irrelevant. Thus, an outcome such as M2, M1, Z3, Z1, Z2 is a winner, whereas M2, Z2, Z1, M1, Z3 is a loser. Actually, the first of these would be declared a winner as soon as M1 was uncovered, and the second would be declared a loser as soon as Z2 was uncovered. However, we show the whole sequence of M’s and Z’s so that we can count outcomes correctly. We then find the probability of winning using an equally likely argument. Specifically, we divide the number of outcomes that are winners by the total number of outcomes. It can be shown that the number of out- comes that are winners is 12, whereas the total number of outcomes is 120. Therefore, the probability of a winner is 12/120 = 0.1.

This calculation, which shows that, on average, 1 out of 10 cards could be win- ners was obviously important for McDonald’s. Actually, this provides only an upper bound on the fraction of cards where a prize was awarded. Many customers threw their cards away without playing the game, and even some of the winners neglected to claim their prizes. So, for example, McDonald’s knew that if they made 50,000 cards where a milk shake was the winning prize, somewhat less than 5000 milk shakes would be given away. Knowing approximately what their expected “losses” would be from win- ning cards, McDonald’s was able to design the game (how many cards of each type to print) so that the expected extra revenue (from customers attracted to the game) would cover the expected losses.

CHAPTER 5 Probability and Probability Distributions

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Copyright 2020 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

5-1 Introduction 1 8 5

5-1 Introduction A key aspect of solving real business problems is dealing appropriately with uncertainty. This involves recognizing explicitly that uncertainty exists and using quantitative methods to model uncertainty. If you want to develop realistic business models, you cannot simply act as if uncertainty doesn’t exist. For example, if you don’t know next month’s demand, you shouldn’t build a model that assumes next month’s demand is a sure 1500 units. This is only wishful thinking. You should instead incorporate demand uncertainty explicitly into your model. To do this, you need to know how to deal quantitatively with uncertainty. This involves probability and probability distributions. We introduce these topics in this chapter and then use them in a number of later chapters.

There are many sources of uncertainty. Demands for products are uncertain, times between arrivals to a supermarket are uncertain, stock price returns are uncertain, changes in interest rates are uncertain, and so on. In many situations, the uncertain quantity— demand, time between arrivals, stock price return, change in interest rate—is a numerical quantity. In the language of probability, it is called a random variable. More formally, a random variable associates a numerical value with each possible random outcome.

Associated with each random variable is a probability distribution that lists all of the possible values of the random variable and their corresponding probabilities. A proba- bility distribution provides very useful information. It not only indicates the possible val- ues of the random variable but it also indicates how likely they are. For example, it is useful to know that the possible demands for a product are, say, 100, 200, 300, and 400, but it is even more useful to know that the probabilities of these four values are, say, 0.1, 0.2, 0.4, and 0.3. This implies, for example, that there is a 70% chance that demand will be at least 300.

It is often useful to summarize the information from a probability distribution with numerical summary measures. These include the mean, variance, and standard deviation. As their names imply, these summary measures are much like the corresponding summary measures in Chapters 2 and 3. However, they are not identical. The summary measures in this chapter are based on probability distributions, not an observed data set. We will use numerical examples to explain the difference between the two—and how they are related.

The purpose of this chapter is to explain the basic concepts and tools necessary to work with probability distributions and their summary measures. The chapter then dis- cusses several important probability distributions, particularly the normal distribution and the binomial distribution, in some detail.

Modeling uncertainty, as we will be doing in the next chapter and later in Chapters 15 and 16, is sometimes difficult, depending on the complexity of the model, and it is easy to get so caught up in the details that you lose sight of the big picture. For this rea- son, the flow chart in Figure 5.1 is useful. (A colored version of this chart is available in the file Modeling Uncertainty Flow Chart.xlsx.) Take a close look at the middle row of this chart. You begin with inputs, some of which are uncertain quantities, you use Excel® formulas to incorporate the logic of the model, and the result is probability distributions of important outputs. Finally, you use this information to make decisions. (The abbreviation EMV stands for expected monetary value. It is discussed extensively in Chapter 6.) The other boxes in the chart deal with implementation issues, particularly with the software you can use to perform the analysis. Study this chart carefully and return to it as you pro- ceed through the next few chapters and Chapters 15 and 16.

Before proceeding, we discuss two terms you often hear in the business world: uncertainty and risk. They are sometimes used interchangeably, but they are not really the same. You typically have no control over uncertainty; it is something that simply exists. A good example is the uncertainty in exchange rates. You cannot be sure what the exchange rate between the U.S. dollar and the euro will be a year from now. All you can try to do is measure this uncertainty with a probability distribution.

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In contrast, risk depends on your position. Even though you don’t know what the exchange rate will be, it makes no difference to you—there is no risk to you—if you have no European investments, you aren’t planning a trip to Europe, and you don’t have to buy or sell anything in Europe. You have risk only when you stand to gain or lose money depending on the eventual exchange rate. Of course, the type of risk you face depends on your position. If you are holding euros in a money market account, you are hoping that euros gain value relative to the dollar. But if you are planning a European vacation, you are hoping that euros lose value relative to the dollar.

Uncertainty and risk are inherent in many of the examples in this book. By learning about probability, you will learn how to measure uncertainty, and you will also learn how to measure the risks involved in various decisions. One important topic you will not learn much about is risk mitigation by various types of hedging. For example, if you know you have to purchase a large quantity of some product from Europe a year from now, you face the risk that the value of the euro could increase dramatically, thus cost- ing you a lot of money. Fortunately, there are ways to hedge this risk, so that if the euro does increase relative to the dollar, your hedge minimizes your losses. Hedging risk is an extremely important topic, and it is practiced daily in the real world, but it is beyond the scope of this book.

5-2 Probability Essentials We begin with a brief discussion of probability. The concept of probability is one that you all encounter in everyday life. When a weather forecaster states that the chance of rain is 70%, he or she is making a probability statement. When a sports commentator states that the odds against the Golden State Warriors winning the NBA Championship are 3 to 1, he or she is also making a probability statement. The concept of probability is quite intuitive. However, the rules of probability are not always as intuitive or easy to master. We examine the most important of these rules in this section.

Assess probability distributions of uncertain inputs:

Decide which inputs are important for the model.

1. Which are known with certainty? 2. Which are uncertain?

Two fundamental approaches:

1. Build an exact probability model that incorporates the rules of probability. (Pros: It is exact and amenable to sensitivity analysis. Cons: It is often difficult mathematically, maybe not even possible.)

2. Build a simulation model. (Pros: It is typically much easier, especially with add-ins like DADM_Tools or @RISK, and extremely versatile. Cons: It is only approximate and runs can be time consuming for complex models).

For simulation models, random values for uncertain inputs are necessary.

1. They can sometimes be generated with built-in Excel functions. This often involves tricks and can be obscure.

2. Add-ins like DADM_Tools or @RISK provide functions that make it much easier.

Examine important outputs.

The result of these formulas should be probability distribution(s) of important output(s). Summarize these probability distributions with (1) histograms (risk profiles), (2) means and standard deviations, (3) percentiles, (4) possibly others.

Model the problem.

Use Excel formulas to relate inputs to important outputs, i.e., enter the business logic.

Make decisions based on this information.

Criterion is usually EMV, but it could be something else, e.g., minimize the probability of losing money.

1. If a lot of historical data is available, find the distribution that best fits the data.

2. Choose a probability distribution (normal? triangular?) that seems reasonable. Add-ins like DADM_Tools or @RISK are helpful for exploring distributions.

3. Gather relevant information, ask experts, and do the best you can.

This is an overview of spreadsheet modeling with uncertainty. The main process is in red. The blue boxes deal with implementation issues.

For simulation models, this can be done “manually” with data tables and built-in functions like AVERAGE, STDEV.S, etc. But add-ins like DADM_Tools or @RISK take care of these bookkeeping details automatically.

Use decision trees, made easier with add-in like DADM_Tools or PrecisionTree, if the number of possible decisions and the number of possible outcomes are not too large.

Figure 5.1 Flow Chart for Modeling Uncertainty

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5-2 Probability Essentials 1 8 7

As the examples in the preceding paragraph illustrate, probabilities are sometimes expressed as percentages or odds. However, these can easily be converted to probabilities on a 0-to-1 scale. If the chance of rain is 70%, then the probability of rain is 0.7. Similarly, if the odds against the Warriors winning are 3 to 1, then the probability of the Warriors winning is 1/4 (or 0.25).

There are only a few probability rules you need to know, and they are discussed in the next few subsections. Surprisingly, these are the only rules you need to know. Probability is not an easy topic, and a more thorough discussion of it would lead to considerable math- ematical complexity, well beyond the level of this book. However, it is all based on the few relatively simple rules discussed next.

5-2a Rule of Complements The simplest probability rule involves the complement of an event. If A is any event, then the complement of A, denoted by A (or in some books by Ac), is the event that A does not occur. For example, if A is the event that the Dow Jones Industrial Average will finish the year at or above the 25,000 mark, then the complement of A is that the Dow will finish the year below 25,000.

If the probability of A is P(A), then the probability of its complement, P(A), is given by Equation (5.1). Equivalently, the probability of an event and the probability of its com- plement sum to 1. For example, if you believe that the probability of the Dow finishing at or above 25,000 is 0.25, then the probability that it will finish the year below 25,000 is 1 − 0.25 = 0.75.

A probability is a number between 0 and 1 that measures the likelihood that some event will occur. An event with probability 0 cannot occur, whereas an event with probability 1 is certain to occur. An event with probability greater than 0 and less than 1 involves uncertainty. The closer its probability is to 1, the more likely it is to occur.

Rule of Complements

P1A2 5 1 2 P1A2 (5.1)

5-2b Addition Rule Events are mutually exclusive if at most one of them can occur. That is, if one of them occurs, then none of the others can occur. For example, consider the following three events involving a company’s annual revenue for the coming year: (1) revenue is less than $1 mil- lion, (2) revenue is at least $1 million but less than $2 million, and (3) revenue is at least $2 million. Clearly, only one of these events can occur. Therefore, they are mutually exclu- sive. They are also exhaustive events, which means that they exhaust all possibilities— one of these three events must occur. Let A1 through An be any n events. Then the addition rule of probability involves the probability that at least one of these events will occur. In general, this probability is quite complex, but it simplifies considerably when the events are mutually exclusive. In this case the probability that at least one of the events will occur is the sum of their individual probabilities, as shown in Equation (5.2). Of course, when the events are mutually exclusive, “at least one” is equivalent to “exactly one.” In addition,

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In a typical application, the events A1 through An are chosen to partition the set of all possible outcomes into a number of mutually exclusive events. For example, in terms of a company’s annual revenue, define A1 as “revenue is less than $1 million,” A2 as “revenue is at least $1 million but less than $2 million,” and A3 as “revenue is at least $2 million.” Then these three events are mutually exclusive and exhaustive. Therefore, their probabilities must sum to 1. Suppose these probabilities are P(A1) = 0.5, P(A2) = 0.3, and P(A3) = 0.2. (Note that these probabilities do sum to 1.) Then the additive rule enables you to calculate other probabilities. For example, the event that revenue is at least $1 million is the event that either A2 or A3 occurs. From the addition rule, its prob- ability is

P1revenue is at least $1 million2 5 P1A22 1 P1A32 5 0.5

Similarly,

P1revenue is less than $2 million2 5 P1A12 1 P1A22 5 0.8

and

P1revenue is less than $1 million or at least $2 million2 5 P1A12 1 P1A32 5 0.7

Again, the addition rule works only for mutually exclusive events. If the events over- lap, the situation is more complex. For example, suppose you are dealt a bridge hand (13 cards from a 52-card deck). Let H, D, C, and S, respectively, be the events that you get at least 5 hearts, at least 5 diamonds, at least 5 clubs, and at least 5 spades. What is the probability that at least one of these four events occurs? It is not the sum of their individual probabilities because they are not mutually exclusive. For example, you could get a hand with 5 hearts and 5 spades. Probabilities such as this one are actually quite difficult to calculate, and we will not pursue them here. Just be aware that the addition rule does not apply unless the events are mutually exclusive.

5-2c Conditional Probability and the Multiplication Rule Probabilities are always assessed relative to the information currently available. As new information becomes available, probabilities can change. For example, if you read that Steph Curry suffered a season-ending injury, your assessment of the probability that the Warriors will win the NBA Championship would obviously change. (It would probably become 0!) A formal way to revise probabilities on the basis of new information is to use conditional probabilities.

Let A and B be any events with probabilities P(A) and P(B). Typically, the probability P(A) is assessed without knowledge of whether B occurs. However, if you are told that B has occurred, then the probability of A might change. The new probability of A is called the conditional probability of A given B, and it is denoted by P(A1B). Note that there is still uncertainty involving the event to the left of the vertical bar in this notation; you do not know whether it will occur. However, there is no uncertainty involving the event to the right of the vertical bar; you know that it has occurred. The conditional probability can be calculated with the following formula.

Addition Rule for Mutually Exclusive Events

P1at least one of A1 through An2 5 P1A12 1 P1A22 1 c 1 P1An2 (5.2)

if the events A1 through An are exhaustive, then the probability is 1 because one of the events is certain to occur.

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5-2 Probability Essentials 1 8 9

The numerator in this formula is the probability that both A and B occur. This proba- bility must be known to find P1AuB2. However, in some applications P1AuB2 and P(B) are known. Then you can multiply both sides of Equation (5.3) by P(B) to obtain the follow- ing multiplication rule for P(A and B).

Multiplication Rule

P1A and B2 5 P1AuB2 P1B2 (5.4)

The conditional probability formula and the multiplication rule are both valid; in fact, they are equivalent. The one you use depends on which probabilities you know and which you want to calculate, as illustrated in Example 5.1.

EXAMPLE

5.1 ASSESSING UNCERTAINTY AT BENDER COMPANY Bender Company supplies contractors with materials for the construction of houses. The company currently has a contract with one of its customers to fill an order by the end of July. However, there is some uncertainty about whether this deadline can be met, due to uncertainty about whether Bender will receive the materials it needs from one of its suppliers by the middle of July. Right now it is July 1. How can the uncertainty in this situation be assessed?

Objective To apply probability rules to calculate the probability that Bender will meet its end-of-July deadline, given the information the company has at the beginning of July.

Solution Let A be the event that Bender meets its end-of-July deadline, and let B be the event that Bender receives the materials from its supplier by the middle of July. The probabilities Bender is best able to assess on July 1 are probably P(B) and P1AuB2. At the beginning of July, Bender might estimate that the chances of getting the materials on time from its supplier are 2 out of 3, so that P(B) = 2/3. Also, thinking ahead, Bender estimates that if it receives the required materials on time, the chances of meeting the end-of-July deadline are 3 out of 4. This is a conditional probability statement that P1AuB2 5 3>4. Then the multiplication rule implies that

P1A and B2 5 P1AuB2P1B2 5 13>42 12>32 5 0.5

That is, there is a fifty–fifty chance that Bender will get its materials on time and meet its end-of-July deadline. This uncertain situation is depicted graphically in the form of a probability tree in Figure 5.2. Note that Bender ini-

tially faces (at the leftmost branch of the tree) the uncertainty of whether event B or its complement will occur. Regardless of whether event B occurs, Bender must next confront the uncertainty regarding event A. This uncertainty is reflected in the set of two pairs of branches in the right half of the tree. Hence, there are four mutually exclusive outcomes regarding the two uncer- tain events, as listed to the right of the tree. Initially, Bender is interested in the first possible outcome, the joint occurrence of

Conditional Probability

P1AuB2 5 P1A and B2

P1B2 (5.3)

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events A and B, the top probability in the figure. Another way to compute this probability is to multiply the probabilities asso- ciated with the branches leading to this outcome, that is, the probability of B times the probability of A given B. As the figure indicates, this is 13/42 12/32, or 0.5.

P(A and B) = (3/4)(2/3)

P(A and B) = (1/4)(2/3)

P(A and B) = (1/5)(1/3)

P(A and B) = (4/5)(1/3)

Figure 5.2 Probability Tree for Bender Example

5-2d Probabilistic Independence A concept that is closely tied to conditional probability is probabilistic independence. You just saw that the probability of an event A can depend on whether another event B has occurred. Typically, the probabilities P(A), P1AuB2, and P1AuB2 are all different, as in Example 5.1. However, there are situations where all of these probabilities are equal. In this case, A and B are probabilistically independent events. This does not mean they are mutually exclusive. Rather, probabilistic independence means that knowledge of one event is of no value when assessing the probability of the other.

There are several other probabilities of interest in this example. First, let B be the complement of B; it is the event that the materials from the supplier do not arrive on time. We know that P1B2 5 1 2 P1B2 5 1>3 from the rule of complements. How- ever, we do not yet know the conditional probability P1AuB2, the probability that Bender will meet its end-of-July deadline, given that it does not receive the materials from the supplier on time. In particular, P1AuB2 is not equal to 1 2 P1AuB2. (Can you see why?) Suppose Bender estimates that the chances of meeting the end-of-July deadline are 1 out of 5 if the materials do not arrive on time, so that P1AuB2 5 1>5. Then a second use of the multiplication rule gives

P1A and B2 5 P1AuB2P1B2 5 11>52 11>32 5 0.0667

In words, there is only 1 chance out of 15 that the materials will not arrive on time and Bender will meet its end-of-July dead- line. This is the third (from the top) probability listed at the right of the tree.

The bottom line for Bender is whether it will meet its end-of-July deadline. After mid-July, this probability is either P1AuB2 5 3>4 or P1AuB2 5 1>5 because by this time, Bender will know whether the materials arrived on time. But on July 1, the relevant probability is P(A)—there is still uncertainty about whether B or B will occur. Fortunately, you can calculate P(A) from the probabilities you already know. The logic is that A consists of the two mutually exclusive events (A and B) and 1A and B2. That is, if A is to occur, it must occur with B or with B. Therefore, the addition rule for mutually exclusive events implies that

P1A2 5 P1A and B2 1 P1A and B2 5 1>2 1 1>15 5 17>30 5 0.5667

The chances are 17 out of 30 that Bender will meet its end-of-July deadline, given the information it has at the beginning of July.

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How can you tell whether events are probabilistically independent? Unfortunately, this issue usually cannot be settled with mathematical arguments. Typically, you need actual data to decide whether independence is reasonable. As a simple example, let A be the event that a family’s first child is male, and let B be the event that its second child is male. Are A and B independent? You could argue that they aren’t independent if you believe, say, that a boy is more likely to be followed by another boy than by a girl. You could argue that they are independent if you believe the chances of the second child being a boy are the same, regardless of the gender of the first child. (Note that neither argument requires that boys and girls are equally likely.)

In any case, the only way to settle the argument is to observe many families with at least two children. If you observe, say, that 55% of all families with first child male also have the second child male, and only 45% of all families with first child male have the sec- ond child female, then you can make a good case that A and B are not independent.

The concept of independence carries over to random variables. Two random variables, X and Y, are independent if any two events, one involving only X and the other involving only Y, are independent. The idea is that knowledge of X is of no help in predicting Y, and vice versa. For example, if X is the amount of rain in Seattle in March and Y is the amount of rain in Seattle in June, it might be realistic to assume that X and Y are independent random variables. March weather probably doesn’t have much effect on June weather. On the other hand, if X and Y are the changes in stock prices of two companies in the same industry from one day to the next, it might not be realistic to assume that X and Y are inde- pendent. The reason is that they might both be subject to the same economic influences.

Note that the multiplication rule applies to events involving independent random vari- ables. For example, if X and Y are independent, then P(X = 10 and Y = 15) equals the prod- uct P(X = 10)P(Y = 15).

It is probably fair to say that most events in the real world are not truly indepen- dent. However, because of the simplified multiplication rule for independent events, many mathematical models assume that events are independent; the math is much easier with this assumption. The question then is whether the results from such a model are believ- able. This depends on how unrealistic the independence assumption really is.

5-2e Equally Likely Events Much of what you know about probability is probably based on situations where outcomes are equally likely. These include flipping coins, throwing dice, drawing balls from urns, and other random mechanisms that are often discussed in introductory probability books. For example, suppose an urn contains 20 red marbles and 10 blue marbles. You plan to randomly select five marbles from the urn, and you are interested, say, in the probability of selecting at least three red marbles. To find this probability, you argue that every possible set of five marbles is equally likely to be chosen. Then you count the number of sets of five marbles that contain at least three red marbles, you count the total number of sets of five marbles that could be selected, and you set the desired probability equal to the ratio of these two counts.

Multiplication Rule for Independent Events

P1A and B2 5 P1A2P1B2 (5.5)

When two events are probabilistically independent, the multiplication rule simpli- fies to Equation (5.5). This follows by substituting P(A) for P1AuB2 in the multiplication rule, which is allowed because of independence. In words, the probability that both events occur is the product of their individual probabilities.

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It is true that many probabilities, particularly in games of chance, can be calculated by using an equally likely argument. It is also true that probabilities calculated in this way satisfy all of the rules of probability. However, many probabilities, especially those in business situations, cannot be calculated by equally likely arguments for the simple reason that the possible outcomes are not equally likely. For example, just because you are able to identify five possible scenarios for a company’s future, there is probably no reason what- soever to conclude that each scenario has probability 1/5.

The bottom line is that there is no need in this book to discuss complex counting rules for equally likely outcomes because most outcomes in the business world are not equally likely.

5-2f Subjective Versus Objective Probabilities We now ask a very basic question: Where do the probabilities in a probability distribution come from? A complete answer to this question could lead to a chapter by itself, so we only briefly discuss the issues involved. There are essentially two distinct ways to assess probabilities, objectively and subjectively. Objective probabilities are those that can be estimated from long-run proportions, whereas subjective probabilities cannot be esti- mated from long-run proportions. Some examples will clarify this distinction.

Consider throwing two dice and observing the sum of the two sides that face up. What is the probability that the sum of these two sides is 7? You might argue as follows. Because there are 6 3 6 5 36 ways the two dice can fall, and because exactly 6 of these result in a sum of 7, the probability of a 7 is 6/36 = 1/6. This is the equally likely argument and it reduces probability to counting.

What if the dice are weighted in some way? Then the equally likely argument is no longer valid. You can, however, toss the dice many times and record the proportion of tosses that result in a sum of 7. This proportion is called a relative frequency.

The relative frequency of an event is the proportion of times the event occurs out of the number of times the random experiment is run.

A famous result called the law of large numbers states that this relative frequency, in the long run, will get closer and closer to the “true” probability of a 7. This is exactly what we mean by an objective probability. It is a probability that can be estimated as the long- run proportion of times an event occurs in a sequence of many identical experiments.

If you are flipping coins, throwing dice, or spinning roulette wheels, objective prob- abilities are certainly relevant. You don’t need a person’s opinion of the probability that a roulette wheel, say, will end up pointing to a red number; you can simply spin it many times and keep track of the proportion of times it points to a red number. However, there are many situations, particularly in business, that cannot be repeated many times—or even more than once—under identical conditions. In these situations objective proba- bilities make no sense (and equally likely arguments usually make no sense either), so you must use subjective probabilities. A subjective probability is one person’s assessment of the likelihood that a certain event will occur. We assume that the person making the assessment uses all of the information available to make the most rational assessment possible.

This definition of subjective probability implies that one person’s assessment of a probability can differ from another person’s assessment of the same probability. For exam- ple, consider the probability that the New England Patriots will win the next Super Bowl. If you ask a casual football observer to assess this probability, you will get one answer, but if you ask a person with a lot of inside information about injuries, team cohesiveness, and so on, you might get a very different answer. Because these probabilities are subjective, people with different information typically assess probabilities in different ways.

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Subjective probabilities are usually relevant for unique, one-time situations. However, most situations are not completely unique; you often have some history to guide you. That is, historical relative frequencies can be factored into subjective probabilities. For exam- ple, suppose a company is about to market a new product. This product might be different in some ways from any products the company has marketed before, but it might also share some features with the company’s previous products. If the company wants to assess the probability that the new product will be a success, it will certainly analyze the unique fea- tures of this product and the current state of the market to obtain a subjective assessment. However, the company will also look at its past successes and failures with reasonably similar products. If the proportion of successes with past products was 20%, say, then this value might be a starting point in the assessment of this product’s probability of success.

All of the “given” probabilities in this chapter and later chapters can be placed some- where on the objective-to-subjective continuum, usually closer to the subjective end. An important implication of this is that these probabilities are not cast in stone; they are usu- ally only educated guesses. Therefore, it is always a good idea to run a sensitivity analysis (especially in Excel, where this is easy to do) to see how “bottom-line” results depend on the given probabilities. Sensitivity analysis is especially important in Chapter 6, when we study decision making under uncertainty.

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 1. In a particular suburb, 30% of the households have

installed electronic security systems. a. If a household is chosen at random from this suburb,

what is the probability that this household has not installed an electronic security system?

b. If two households are chosen at random from this sub- urb, what is the probability that neither has installed an electronic security system?

2. Several major automobile producers are competing to have the largest market share for sport utility vehicles (SUVs) in the coming quarter. A professional automo- bile market analyst assesses that the odds of General Motors not being the market leader are 6 to 1. The odds against Toyota and Ford having the largest market share in the coming quarter are similarly assessed to be 12 to 5 and 8 to 3, respectively. a. Find the probability that General Motors will have the

largest market share for SUVs in the coming quarter. b. Find the probability that Toyota will have the largest

market share for SUVs in the coming quarter. c. Find the probability that Ford will have the largest

market share for SUVs in the coming quarter. d. Find the probability that some other automobile man-

ufacturer will have the largest market share for SUVs in the coming quarter.

3. The publisher of a popular financial periodical has decided to undertake a campaign in an effort to attract new subscribers. Market research analysts in this

company believe that there is a 1 in 4 chance that the increase in the number of new subscriptions resulting from this campaign will be less than 3000, and there is a 1 in 3 chance that the increase in the number of new subscriptions resulting from this campaign will be between 3000 and 5000. What is the probability that the increase in the number of new subscriptions result- ing from this campaign will be less than 3000 or more than 5000?

4. Suppose that you draw a single card from a standard deck of 52 playing cards. a. What is the probability that this card is a diamond or

club? b. What is the probability that this card is not a 4? c. Given that this card is a black card, what is the proba-

bility that it is a spade? d. Let E1 be the event that this card is a black card. Let

E2 be the event that this card is a spade. Are E1 and E2 independent events? Why or why not?

e. Let E3 be the event that this card is a heart. Let E4 be the event that this card is a 3. Are E3 and E4 indepen- dent events? Why or why not?

Level B 5. In a large accounting firm, the proportion of accountants

with MBA degrees and at least five years of professional experience is 75% as large as the proportion of accoun- tants with no MBA degree and less than five years of pro- fessional experience. Furthermore, 35% of the accountants in this firm have MBA degrees, and 45% have fewer than five years of professional experience. If one of the firm’s accountants is selected at random, what is the probability that this accountant has an MBA degree or at least five years of professional experience, but not both?

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6. A local beer producer sells two types of beer, a regular brand and a light brand with 30% fewer calories. The company’s marketing department wants to verify that its traditional approach of appealing to local white-collar workers with light beer commercials and appealing to local blue-collar workers with regular beer commercials is indeed a good strategy. A randomly selected group of 400 local workers are questioned about their beer- drinking preferences, and the data in the file P05_06. xlsx are obtained. a. If a blue-collar worker is chosen at random from this

group, what is the probability that he/she prefers light beer (to regular beer or no beer at all)?

b. If a white-collar worker is chosen at random from this group, what is the probability that he/she prefers light beer (to regular beer or no beer at all)?

c. If you restrict your attention to workers who like to drink beer, what is the probability that a randomly selected blue-collar worker prefers to drink light beer?

d. If you restrict your attention to workers who like to drink beer, what is the probability that a randomly selected white-collar worker prefers to drink light beer?

e. Does the company’s marketing strategy appear to be appropriate? Explain why or why not.

7. Suppose that two dice are tossed. For each die, it is equally likely that 1, 2, 3, 4, 5, or 6 dots will turn up. Let S be the sum of the two dice. a. What is the probability that S will be 5 or 7? b. What is the probability that S will be some number

other than 4 or 8? c. Let E1 be the event that the first die shows a 3. Let E2 be

the event that S is 6. Are E1 and E2 independent events? d. Again, let E1 be the event that the first die shows a 3.

Let E3 be the event that S is 7. Are E1 and E3 indepen- dent events?

e. Given that S is 7, what is the probability that the first die showed 4 dots?

f. Given that the first die shows a 3, what is the probabil- ity that S is an even number?

5-3 Probability Distribution of a Random Variable We now discuss one of the most important concepts in this book, the probability distribu- tion of a random variable.

There are really two types of random variables: discrete and continuous. A discrete random variable has only a finite number of possible values, whereas a continuous ran- dom variable has a continuum of possible values.1 Usually a discrete distribution results from a count, whereas a continuous distribution results from a measurement. For example, the number of children in a family is clearly discrete, whereas the amount of rain this year in San Francisco is clearly continuous.

Concept of Probability Distribution

A probability distribution describes the uncertainty of a numerical outcome. It is not based, at least not directly, on a data set of the type discussed in the previ- ous chapters. Instead, it is a list of all possible outcomes and their corresponding probabilities.

Fundamental Insight

Mathematically, there is an important difference between discrete and continuous probability distributions. Specifically, a proper treatment of continuous distributions, anal- ogous to the treatment we provide in this section, requires calculus—which we do not pre- sume for this book. Therefore, we discuss only discrete distributions in this section. Later in the book, we often use continuous distributions, particularly the bell-shaped normal distribution, but we simply state their properties without deriving them mathematically.

1 Actually, a more rigorous discussion allows a discrete random variable to have an infinite number of possible values, such as all positive integers.

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5-3 Probability Distribution of a Random Variable 1 9 5

The essential properties of a discrete random variable and its associated probability dis- tribution are quite simple. We discuss them in general and then analyze a numerical example.

Let X be a random variable. To specify the probability distribution of X, we need to specify its possible values and their probabilities. We assume that there are k possible values, denoted x1, x2, …, xk. The probability of a typical value xi is denoted in one of two ways, either P(X = xi) or p(xi). The first is a reminder that this probability involves the ran- dom variable X, whereas the second is a shorthand notation. Probability distributions must satisfy two criteria: (1) the probabilities must be nonnegative, and (2) they must sum to 1. In symbols, we must have

i5 1 p1xi2 5 1, p1xi2 $ 0

This is basically all there is to it: a list of possible values and a list of associated prob- abilities that sum to 1. It is also sometimes useful to calculate cumulative probabilities. A cumulative probability is the probability that the random variable is less than or equal to some particular value. For example, assume that 10, 20, 30, and 40 are the possible values of a random variable X, with corresponding probabilities 0.15, 0.25, 0.35, and 0.25. Then a typical cumulative probability is P1X # 302. From the addition rule it can be calculated as

P1X # 302 5 P1X 5 102 1 P1X 5 202 1 P1X 5 302 5 0.75

The point is that the cumulative probabilities are determined by the individual probabilities.

5-3a Summary Measures of a Probability Distribution It is often convenient to summarize a probability distribution, discrete or continuous, with two or three well-chosen numbers. The first of these is the mean, often denoted m. It is also called the expected value of X and denoted E(X) (for expected X). The mean of a dis- crete distribution is a weighted sum of the possible values, weighted by their probabilities, as shown in Equation (5.6). In much the same way that an average of a set of numbers indicates “central location,” the mean indicates the “center” of the probability distribution.

Usually, capital letters toward the end of the alphabet, such as X, Y, and Z, are used to denote random variables.

A discrete probability distribution is a set of possible values and a corresponding set of probabilities that sum to 1.

Mean of a Probability Distribution, m

m 5 E1X2 5 ak

i5 1 xi p1xi2 (5.6)

To measure the variability in a distribution, we calculate its variance or standard devi- ation. The variance of a discrete distribution, denoted by s2 or Var1X2, is a weighted sum of the squared deviations of the possible values from the mean, where the weights are again the probabilities. This is shown in Equation (5.7). As in Chapter 2, the variance is expressed in the square of the units of X, such as dollars squared. Therefore, a more nat- ural measure of variability is the standard deviation, denoted by s or Stdev1X2. It is the square root of the variance, as indicated by Equation (5.8).

Variance of a Probability Distribution, s2

s2 5 Var1X2 5 ak

i 3xi 2 E1X2 42 p1xi2 (5.7)

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Equation (5.7) is useful for understanding variance as a weighted average of squared deviations from the mean. However, the following is an equivalent formula for variance and is somewhat easier to implement in Excel. (It can be derived with straightforward algebra.) In words, you find the weighted average of the squared values, weighted by their probabilities, and then subtract the square of the mean.

Standard Deviation of a Probability Distribution, s

s 5 Stdev1X2 5 !Var1X2 (5.8)

Variance (computing formula)

s2 5 ak

i5 1 xi

2 p1xi2 2 m2 (5.9)

We now consider a typical example.

EXAMPLE

5.2 MARKET RETURN SCENARIOS FOR THE NATIONAL ECONOMY An investor is concerned with the market return for the coming year, where the market return is defined as the percent- age gain (or loss, if negative) over the year. The investor believes there are five possible scenarios for the national econ- omy in the coming year: rapid expansion, moderate expansion, no growth, moderate contraction, and serious contraction. Furthermore, she has used all of the information avail- able to her to estimate that the market returns for these scenarios are, respectively, 23%, 18%, 15%, 9%, and 3%. That is, the possible returns vary from a high of 23% to a low of 3%. Also, she has assessed that the probabilities of these outcomes are 0.12, 0.40, 0.25, 0.15, and 0.08. Use this information to describe the probability distribution of the market return.

Objective To compute the mean, variance, and standard deviation of the probability distribution of the market return for the coming year.

Solution To make the connection between the general notation and this particular example, let X denote the market return for the coming year. Then each possible economic scenario leads to a possible value of X. For example, the first possible value is x1 5 23%, and its probability is p1x12 5 0.12. These values and probabilities appear in columns B and C of Figure 5.3. (See the file Market Return Finished.xlsx.) Note that the five probabilities sum to 1, as they should. This probability distribution implies, for example, that the probability of a market return at least as large as 18% is 0.12 1 0.40 5 0.52 because it could occur as a result of rapid or moderate expansion of the economy. Similarly, the probability that the market return is 9% or less is 0.15 1 0.08 5 0.23 because this could occur as a result of moderate or serious contraction of the economy.

The summary measures of this probability distribution appear in the range B11:B13. They can be calculated with the fol- lowing steps. (Note that the formulas make use of the range names listed in the figure.)

In reality, there is a continuum of possible returns. The assumption of only five possible returns is clearly an approximation to reality, but such an approxi- mation is often useful.

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Procedure for Calculating Summary Measures 1. Mean return. Calculate the mean return in cell B11 with the formula

5SUMPRODUCT(Market_return,Probability)

Figure 5.3 Probability Distribution of Market Returns

1 2 3 4 5 6 7 8 9

10 11 12 13

A B C D E F G Mean, variance, and standard deviation of the market return �an�e names �sed

Market_return =$C$4:$C$8 Economic outcome Probability Market return Sq dev from mean Mean =$B$11 Rapid Expansion 0.005929 Probability =$B$4:$B$8 Moderate Expansion 0.000729

0.000009 Sq_dev_from_mean =$D$4:$D$8

No Growth Stdev Moderate Contraction 0.003969 Variance =$B$12

=$B$13

Serious Contraction

Summary measures of return Mean Variance 0.002811

15.3%

5.3% 0.002811 Quick alternative formula

5.3%Stdev

0.015129

0.12 0.23 0.40 0.18

0.150.25 0.15 0.09

0.030.08

Excel’s SUMPRODUCT Function

Excel’s SUMPRODUCT function is a gem, and you should use it whenever possible. It takes (at least) two argu- ments, which must be ranges of exactly the same size and shape. It sums the products of the values in these rang- es. For example, =SUMPRODUCT 1A1:A3,B1:B32 is equivalent to the formula =A1*B1 1 A2*B2 1 A3*B3. If the ranges contain only a few cells, there isn’t much advantage to using SUMPRODUCT, but when the ranges are large, such as A1:A100 and B1:B100, SUMPRODUCT is the only viable option.

Excel Tip

This formula illustrates the general rule in Equation (5.6): The mean is the sum of products of possible values and probabilities.

2. Squared deviations. To get ready to compute the variance from Equation (5.7), calculate the squared deviations from the mean by entering the formula

5(C4-Mean)^2

in cell D4 and copying it down through cell D8. 3. Variance. Calculate the variance of the market return in cell B12 with the formula

5SUMPRODUCT(Sq_dev_from_mean,Probability)

This illustrates the general formula for variance in Equation (5.7). The variance is always a sum of products of squared deviations from the mean and probabilities. Alternatively, you can skip the calculation of the squared deviations from the mean and use Equation (5.9) directly. This is done in cell C12 with the formula

5SUMPRODUCT(Market_return,Market_return,Probability)-Mean^2

By entering the Market_return range twice in this SUMPRODUCT formula, you get the squares. From now on, we will use this simplified formula for variance and dispense with squared deviations from the mean. But regardless of how it is calculated, you should remember the essence of variance: It is a weighted average of squared deviations from the mean.

4. Standard deviation. Calculate the standard deviation of the market return in cell B13 with the formula

5SQRT(Variance)

As always, range names are not required, but they make the Excel formulas easier to read. You can use them or omit them, as you wish.

5-3 Probability Distribution of a Random Variable 1 9 7

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You can see that the mean return is 15.3% and the standard deviation is 5.3%. What do these measures really mean? First, the mean, or expected, return does not imply that the most likely return is 15.3%, nor is this the value that the investor “expects” to occur. In fact, the value 15.3% is not even a possible market return, at least not according to the model. You can understand these measures better in terms of long-run averages. Specifically, if you could imagine the coming year being repeated many times, each time using the probability distribution in columns B and C to generate a market return, then the average of these market returns would be close to 15.3%, and their standard deviation—calculated as in Chapter 2—would be close to 5.3%.

Before leaving this section, we emphasize a key point, a point that is easy to forget with all the details. The whole point of discussing probability and probability distributions, especially in the context of business problems, is that uncertainty is often a key factor, and you cannot simply ignore it. The mean return in this example is 15.3%. However, it would be far from realistic to treat the actual return as a sure 15.3%, with no uncertainty. If you did this, you would be ignoring the uncertainty completely, and it is often the uncertainty that makes business problems interesting—and adds risk. Therefore, to model such problems in a realistic way, you must deal with probability and probability distributions.

2 This section is somewhat advanced and can be omitted. It won’t be used in the rest of the book.

Conditional Mean Formula

E1X2 5 ak

i5 1 Ei1X2pi (5.10)

Conditional Variance Formula

Var 1X2 5 ak

i5 1 5Vari1X2 1 3Ei1X2 426pi 2 3E 1X2 42 (5.11)

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5-3b Conditional Mean and Variance2

There are many situations where the mean and variance of a random variable depend on some external event. In this case, you can condition on the outcome of the external event to find the overall mean and variance (or standard deviation) of the random variable.

It is best to motivate this with an example. Consider the random variable X, repre- senting the percentage change in the price of stock A from now to a year from now. This change is driven partly by circumstances specific to company A, but it is also driven partly by the economy as a whole. In this case, the outcome of the economy is the external event. Let’s assume that the economy in the coming year will be awful, stable, or great with probabilities 0.20, 0.50, and 0.30, respectively. (Of course, these are subjective probabili- ties.) In addition, we make the following assumptions.

• Given that the economy is awful, the mean and standard deviation of X are 220% and 30%.

• Given that the economy is stable, the mean and standard deviation of X are 5% and 20%.

• Given that the economy is great, the mean and standard deviation of X are 25% and 15%.

Each of these statements is a statement about X conditional upon the economy. What can we say about the unconditional mean and standard deviation of X? That is, what are the mean and standard deviation of X before we learn the state of the economy? The answers come from Equations (5.10) and (5.11). In the context of the example, pi is the probability of economy state i, and Ei1X2 and Vari1X2 are the mean and variance of X, given that econ- omy state i occurs.

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5-3 Probability Distribution of a Random Variable 1 9 9

In the example, the mean percentage change in the price of stock A, from Equation (5.10), is

E1X2 5 0.21220%2 1 0.515%2 1 0.3125%2 5 6%

To calculate the standard deviation of X, first use Equation (5.11) to calculate the variance, and then take its square root. The variance is

Var1X2 5 50.23 130%22 1 1220%224 1 0.53 120%22 1 15%224

1 0.33 115%22 1 125%224 6 2 16%22 5 0.06915

Taking the square root gives

Stdev1X2 5 !0.06915 5 26.30%

Of course, these calculations can be done easily in Excel. See the file Stock Price and Economy Finished.xlsx for details.

The point of this example is that it is often easier to assess the uncertainty of some random variable X by conditioning on every possible outcome of some external event like the economy. However, before that outcome is known, the relevant mean and standard deviation of X are those calculated from Equations (5.10) and (5.11). In this particular example, before you know the state of the economy, the relevant mean and standard devia- tion of the change in the price of stock A are 6% and 26.3%, respectively.

the random variable X, the number of months from now it will take to complete this project: 2, 2.5, 3, and 3.5. The manager currently thinks that the probabilities of these four possibilities are in the ratio 1 to 2 to 4 to 2. That is, X 5 2.5 is twice as likely as X 5 2, X 5 3 is twice as likely as X 5 2.5, and X 5 3.5 is half as likely as X 5 3. a. Find the probability distribution of X. b. What is the probability that this project will be com-

pleted in less than three months from now? c. What is the probability that this project will not be

completed on time? d. What is the expected completion time (in months) of

this project from now? e. How much variability (in months) exists around the

expected value you found in part d?

Level B 11. The National Football League playoffs are just about to

begin. Because of their great record in the regular sea- son, the Steelers get a bye in the first week of the play- offs. In the second week, they will play the winner of the game between the Ravens and the Patriots. A football expert estimates that the Ravens will beat the Patriots with probability 0.45. This same expert estimates that if the Steelers play the Ravens, the mean and standard deviation of the point spread (Steelers points minus Ravens points) will be 6.5 and 10.5, whereas if the Steelers play the Patriots, the mean and standard devi- ation of the point spread (Steelers points minus Patriots points) will be 3.5 and 12.5. Find the mean and standard deviation of the point spread (Steelers points minus their opponent’s points) in the Steelers game.

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 8. Consider a random variable with the following proba-

bility distribution: P1X 5 02 5 0.1, P1X 5 12 5 0.2, P1X 5 22 5 0.3, P1X 5 32 5 0.3, and P1X 5 42 5 0.1. a. Find P1X # 22. b. Find P11 6 X # 32. c. Find P1X 7 02. d. Find P1X 7 3|X 7 22. e. Find the expected value of X. f. Find the standard deviation of X.

9. A study has shown that the probability distribu- tion of X, the number of customers in line (including the one being served, if any) at a checkout counter in a department store, is given by P1X 5 02 5 0.25, P1X 5 12 5 0.25, P1X 5 22 5 0.20, P1X 5 32 5 0.20, and P1 $ 42 5 0.10. Consider a newly arriving cus- tomer to the checkout line. a. What is the probability that this customer will not

have to wait behind anyone? b. What is the probability that this customer will have to

wait behind at least one customer? c. On average, the newly arriving customer will have to

wait behind how many other customers? 10. A construction company has to complete a project no

later than three months from now or there will be sig- nificant cost overruns. The manager of the construction company believes that there are four possible values for

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5-4 The Normal Distribution In the rest of this chapter, we discuss specific probability distributions, starting with the single most important distribution in statistics, the normal distribution. It is a continuous distribution and is the basis of the familiar symmetric bell-shaped curve. Any particular normal distribution is specified by its mean and standard deviation. By changing the mean, the normal curve shifts to the right or left. By changing the standard deviation, the curve becomes more or less spread out. Therefore, there are really many normal distributions, not just a single one. We say that the normal distribution is a two-parameter family, where the two parameters are the mean and the standard deviation.

5-4a Continuous Distributions and Density Functions We first take a moment to discuss continuous probability distributions in general. In the previous section we discussed discrete distributions, characterized by a list of possible values and their probabilities. The same basic idea holds for continuous distributions such as the normal distribution, but the mathematics is more complex. Now instead of a list of possible values, there is a continuum of possible values, such as all values between 0 and 100 or all values greater than 0. Instead of assigning probabilities to each individual value in the continuum, the total probability of 1 is spread over this continuum. The key to this spreading is called a density function, which acts like a histogram. The higher the value of the density function, the more likely this region of the continuum is.

12. The “house edge” in any game of chance is defined as

E1player’s loss on a bet2 Size of player’s loss on a bet

For example, if a player wins $10 with probability 0.48 and loses $10 with probability 0.52 on any bet, the house edge is

2 31010.482 2 1010.522 4 10

5 0.04

Give an interpretation to the house edge that relates to how much money the house is likely to win on average. Which do you think has a larger house edge: roulette or sports gambling? Why?

A density function, usually denoted by f1x2, specifies the probability distribu- tion of a continuous random variable X. The higher f1x2 is, the more likely x is. Also, the total area between the graph of f1x2 and the horizontal axis, which represents the total probability, is equal to 1. Finally, f1x2 is nonnegative for all possible values of X.

As an example, consider the density function shown in Figure 5.4. (This is not a nor- mal density function.) It indicates that all values in the continuum from 25 to 100 are pos- sible, but that the values near 70 are most likely. (This density function might correspond to scores on an exam.) More specifically, because the height of the density at 70 is approx- imately twice the height of the curve at 84 or 53, a value near 70 is approximately twice as likely as a value near 84 or a value near 53. In this sense, the height of the density function indicates relative likelihoods.

Probabilities are found from a density function as areas under the curve. For exam- ple, the area of the designated region in Figure 5.5 represents the probability of a score between 65 and 75. Also, the area under the entire curve is 1 because the total probability of all possible values is always 1. Fortunately, Excel functions have been developed to find these areas—without the need for bulky tables. We take advantage of these Excel func- tions in the rest of this chapter and in later chapters.

For continuous distributions, probabilities are areas under the density function. These probabilities can often be calculated with Excel functions.

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5-4 The Normal Distribution 2 0 1

Figure 5.4 A Skewed Density Function

3.5

2.5

1.5

0.5

0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99

As in the previous chapter, the mean is a measure of central tendency of the distribu- tion, and the standard deviation (or variance) measures the variability of the distribution. We will list their values for the normal distribution and any other continuous distributions where we need them. By the way, the mean for the (nonnormal) density in Figure 5.4 is slightly less than 70—it is always to the left of the peak for a left-skewed distribution and to the right of the peak for a right-skewed distribution—and the standard deviation is approximately 15.

5-4b The Normal Density Function The normal distribution is a continuous distribution with possible values ranging over the entire number line—from “minus infinity” to “plus infinity.” However, only a rela- tively small range has much chance of occurring. The normal density function is actually quite complex, in spite of its “nice” bell-shaped appearance. For the sake of completeness, we list the formula for the normal density function in Equation (5.12). Here, m and s are the mean and standard deviation of the distribution.

Figure 5.5 Probability as the Area Under the Density

3.5

2.5

1.5

0.5

0 0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99

Area under curve is probability of being between 65 and 75

Normal Density Function

f 1x2 5 1

!2ps e2 1x 2 m22/12s22 for 2 ` 6 x 6 1 ` (5.12)

The curves in Figure 5.6 illustrate several normal density functions for different values of m and s. The mean m can be any number: negative, positive, or zero. As you can see, the effect of increasing or decreasing the mean m is to shift the curve to the right or the left. On the other hand, the standard deviation s must be a positive number. It controls the

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spread of the normal curve. When s is small, the curve is more peaked; when s is large, the curve is more spread out. For shorthand, we use the notation N1m, s2 to refer to the normal distribution with mean m and standard deviation s. For example, N122, 12 refers to the normal distribution with mean 22 and standard deviation 1.

Figure 5.6 Several Normal Density Functions

5-4c Standardizing: Z-Values There are infinitely many normal distributions, one for each pair m and s. We single out one of these for special attention, the standard normal distribution. The standard normal distribution has mean 0 and standard deviation 1, so we denote it by N10, 12 . It is also commonly referred to as the Z distribution. Suppose the random variable X is normally distributed with mean m and standard deviation s. We define the random variable Z by Equation (5.13). This operation is called standardizing. That is, to standardize a variable, you subtract its mean and then divide the difference by the standard deviation. When X is normally distributed, the standardized variable is N10, 12.

Why the Normal Distribution?

The normal density in Equation (5.12) is certainly not very intuitive, so why is the normal distribution the basis for so much of statistical theory? One reason is practical. Many histograms based on real data resemble the bell-shaped nor- mal curve to a remarkable extent. Granted, not all histograms are symmetric and bell shaped, but a surprising number are. Another reason is theoretical. In spite of the complexity of Equation (5.12), the normal distribution has many appealing properties that have enabled researchers to build the rich statistical theory that finds widespread use in business, the sciences, and other fields.

Fundamental Insight

Standardizing a Normal Random Variable

Z 5 X 2 m

s (5.13)

One reason for standardizing is to measure variables with different means and/or stan- dard deviations on a single scale. For example, suppose several sections of a college course are taught by different instructors. Because of differences in teaching methods and grading

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5-4 The Normal Distribution 2 0 3

procedures, the distributions of scores in these sections might differ. However, if each instruc- tor calculates his or her mean and standard deviation and then calculates a Z-value for each student, the distributions of the Z-values should be approximately the same in each section.

It is easy to interpret a Z-value. It is the number of standard deviations to the right or the left of the mean. If Z is positive, the original value (in this case, the original score) is to the right of the mean; if Z is negative, the original score is to the left of the mean. For example, if the Z-value for some student is 2, then this student’s score is two standard deviations above the mean. If the Z-value for another student is 20.5, then this student’s score is half a standard deviation below the mean. We illustrate Z-values in Example 5.1.

EXAMPLE

5.3 STANDARDIZING RETURNS FROM MUTUAL FUNDS The annual returns for 30 mutual funds appear in Figure 5.7. (See the file Standardizing Finished.xlsx.) Find and interpret the Z-values of these returns.

Objective To use Excel to standardize annual returns of various mutual funds.

A B C D E F

Standardizing mutual fund returns

Summary statistics from values below

Mean

Stdev

Fund Annual return Z value

0.091

0.047

0.000

1.000

0.007

0.080

0.082

0.123

0.022

0.054

0.094

–1.8047

–0.2363

Mean

Stdev

=$B$4

=$B$5

–0.1934

0.6875

–1.4824

–0.7949

0.0645

Range names used

0.078 –0.279336 29

0.066 –0.537137 30

Figure 5.7 Mutual Fund Returns and Z-Values

Solution The 30 annual returns appear in column B of Figure 5.7. Their mean and standard deviation are calculated in cells B4 and B5 with the AVERAGE and STDEV.S functions. The corresponding Z-values are calculated in column C by entering the formula

5(B8-Mean)/Stdev

in cell C8 and copying it down column C. (Note that Mean and Stdev are range names for cells B4 and B5.) The Z-values in Figure 5.7 range from a low of 21.80 to a high of 2.19. Specifically, the return for stock 1 is about 1.80

standard deviations below the mean, whereas the return for fund 17 is about 2.19 standard deviations above the mean. As you will see shortly, these values are typical: Z-values are usually in the range from 22 to 12 and values beyond −3 or 13 are very uncommon. (Recall the empirical rules for interpreting standard deviation discussed in Chapter 2.) Also, the Z-values automatically have mean 0 and standard deviation 1, as you can see in cells C4 and C5 by using the AVERAGE and STDEV.S functions on the Z-values in column C.

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2 0 4 C H A P T E R 5 P r o b a b i l i t y a n d P r o b a b i l i t y D i s t r i b u t i o n s

5-4d Normal Tables and Z-Values3

A common use for Z-values and the standard normal distribution is in calculating proba- bilities and percentiles by the traditional method. This method is based on a table of the standard normal distribution found in many statistics textbooks. Such a table is given in Figure 5.8. The body of the table contains probabilities. (The entire body of this table was generated by the single copyable formula shown.) The left and top margins contain possible values. Specifically, suppose you want to find the probability that a standard normal random variable is less than 1.35. You locate 1.3 along the left and 0.05—the second decimal in 1.35—along the top, and then read into the table to find the probability 0.9115. In words, the probability is about 0.91 that a standard normal random variable is less than 1.35.

Figure 5.8 Table of Normal Probabilities

z 0.00 0.01 0.02 0.03 0.05 0.06 0.07 0.08 0.09 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133

0.8159 0.8186 0.8212 0.8238 0.8264 0.8365 0.8389 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890 2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936 2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964 2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974 2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981 2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986 3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990 3.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.9993 3.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.9995 3.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9997 3.4 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9998

A B C D E G H I J K

0.9

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

=NORM.S.DIST($A11+F$1, TRUE) NORM.S.DIST(z, cumulative)

0.04 F

3 If you intend to rely on Excel functions for normal calculations (and we hope you do!), you can skip this subsection.

09953_ch05_ptg01_183-241.indd 204 04/03/19 12:25 PM

5-4 The Normal Distribution 2 0 5

Alternatively, if you are given a probability, you can use the table to find the value with this much probability to the left of it under the standard normal curve. This is called a percentile calculation. For example, if the probability is 0.75, you can find the 75th percentile by locating the probability in the table closest to 0.75 and then reading to the left and up. With interpolation, the required value is approximately 0.675. In words, the probability of being to the left of 0.675 under the standard normal curve is approximately 0.75.

There are some obvious drawbacks to using the standard normal table for probability calculations. The first is that there are holes in the table—interpolation is often necessary. A second drawback is that the standard normal table takes different forms in different textbooks. These differences are rather minor, but they can easily cause confusion. Finally, the table requires you to perform calculations, where errors are easy to make.

The Excel functions discussed next make the whole procedure much easier and less error-prone.

5-4e Normal Calculations in Excel Two types of calculations are typically made with normal distributions: finding probabil- ities and finding percentiles. Excel has functions for both of these. The functions used for normal probability calculations are NORM.DIST and NORM.S.DIST. The main dif- ference between these is that the one with the “S” (for standard) applies only to N10, 12 calculations, whereas NORM.DIST applies to any normal distribution. On the other hand, percentile calculations that take a probability and return a value are often called inverse calculations. Therefore, the Excel functions for these are named NORM.INV and NORM.S.INV. Again, the “S” in the second of these indicates that it applies to the standard normal distribution.

The NORM.DIST and NORM.S.DIST functions return left-tail probabilities, such as the probability that a normally distributed variable is less than 35. The syntax for these functions is

5NORM.DIST(x,m,s,TRUE)

and

5NORM.S.DIST (x,TRUE)

Here, x is a number you supply, and m and s are the mean and standard deviation of the normal distribution. The last argument in the NORM.DIST function, TRUE (or 1), is used to obtain the cumulative normal probability, the type usually required. (This TRUE/FALSE argument is a nuisance to remember, but it is necessary. If the last argument is FALSE, or 0, the function returns the height of the density at x, which is not usually what you want.) The NORM.S.DIST function takes only two arguments (because m and s are known to be 0 and 1), so it is easier to use—when it applies. (Note that the “old” NORMSDIST function doesn’t require the last TRUE/FALSE argument, but the newer NORM.S.DIST function, introduced in Excel 2010, does.)

The NORM.INV and NORM.S.INV functions return values for user-supplied probabilities. For example, if you supply the probability 0.95, these functions return the 95 th percentile. Their syntax is

=NORM.INV(p,m,s)

and

=NORM.S.INV(p)

where p is a probability you supply. These are analogous to the NORM.DIST and NORM.S.DIST functions except there is no final TRUE/FALSE argument in the “INV” functions.

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2 0 6 C H A P T E R 5 P r o b a b i l i t y a n d P r o b a b i l i t y D i s t r i b u t i o n s

We illustrate these Excel functions in the following example.4

Old Normal Functions

Prior to Excel 2010, the normal functions were NORMDIST, NORMSDIST, NORMINV, and NORMSINV. They are almost the same as the new functions, the ones with periods in their names, except that NORMSDIST does not take a last TRUE/FALSE argument. Microsoft kept these old functions for backward compatibility, so you can continue to use them if you prefer.

Excel Tip

Probability and Percentile Calculations

There are two basic types of calculations involving probability distributions, nor- mal or otherwise. In a probability calculation, you provide a possible value, and you ask for the probability of being less than or equal to this value. In a percentile calculation, you provide a probability, and you ask for the value that has this probability to the left of it. Excel’s statistical functions use DIST functions to perform probability calculations and INV (for inverse) functions to perform per- centile calculations.

Fundamental Insight

EXAMPLE

5.4 NORMAL CALCULATIONS IN EXCEL Use Excel to calculate the following probabilities and percentiles for the standard normal distribution: (a) P1Z 6 2 22, (b) P1Z 7 12 , (c) P120.4 6 Z 6 1.62 , (d) the 5th percentile, (e) the 75th percentile, and (f) the 99th percentile. Then for the N175, 82 distribution, find the following probabilities and percentiles: (a) P1X 6 702, (b) P1X 7 732, (c) P175 6 X 6 852, (d) the 5th percentile, (e) the 60th percentile, and (f) the 97th percentile.

Objective To calculate probabilities and percentiles for standard normal and general normal distributions in Excel.

Solution The solution appears in Figure 5.9. (See the file Normal Calculations Finished.xlsx.) The N10, 12 calculations are in rows 7 through 14; the N175, 82 calculations are in rows 23 through 30. For your convenience, the formulas used in column B are spelled out in column D (as labels). Note that the standard normal calculations use the normal functions with the “S” in the middle; the rest use the normal functions without the “S”—and require more arguments.

Note the following for normal probability calculations:

• For “less than” probabilities, use NORM.DIST or NORM.S.DIST directly. (See rows 7 and 23.)

• For “greater than” probabilities, subtract the NORM.DIST or NORM.S.DIST function from 1. (See rows 8 and 24.) • For “between” probabilities, subtract the two NORM.DIST or NORM.S.DIST functions. For example, in row 9 the proba-

bility of being between 20.4 and 1.6 is the probability of being less than 1.6 minus the probability of being less than 20.4.

4 Actually, we already illustrated the NORM.S.DIST function; it was used to create the body of Figure 5.8. In other words, you can use it to build your own normal probability table.

09953_ch05_ptg01_183-241.indd 206 04/03/19 12:25 PM

The percentile calculations are even more straightforward. In most percentile problems you want to find the value with a cer- tain probability to the left of it. In this case you use the NORM.INV or NORM.S.INV function with the specified probability as the first argument. See rows 12 through 14 and 28 through 30.

Note that when you are calculating probabilities for any continuous distribution, including the normal distribution, there is no need to distinguish between “less than” and “less than or equal to” events, or between “greater than” and “greater than or equal to” events. The reason is that there is no positive probability of being equal to any particular value. However, as you will see when we discuss the binomial distribution, this is not true of discrete distributions.

There are a couple of variations of percentile calculations. First, suppose you want the value with probability 0.05 to the right of it. This is the same as the value with probability 0.95 to the left of it, so you use NORM.INV or NORM.S.INV with probability argument 0.95. For example, the value with probability 0.4 to the right of it in the N175, 82 distribution is the 60th percentile, 77.027. (See cell B29 in Figure 5.6.)

As a second variation, suppose you want to find an interval of the form 2x to x, for some positive number x, with (1) probability 0.025 to the left of 2x, (2) probability 0.025 to the right of x, and (3) probability 0.95 between 2x and x. This is a very common problem in statistical inference. In general, you want a probability (such as 0.95) to be in the middle of the interval so that half of the remaining probability (0.025) is in each of the tails. (See Figure 5.10.) Then the required x

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

A B C D E F G H I

Range Probability Less than –2 0.0228 =NORM.S.DIST(–2,TRUE) Greater than 1 0.1587 =1-NORM.S.DIST(1,TRUE) Between –0.4 and 1.6 0.6006 =NORM.S.DIST(1.6,TRUE)-NORM.S.DIST(–0.4,TRUE)

5th –1.645 =NORM.S.INV(0.05)

=NORM.S.INV(0.99) =NORM.S.INV(0.75)0.67475th

2.32699th

Range names used: MeanMean StdevStdev

Range Probability Less than 70 0.2660 =NORM.DIST(70,Mean,Stdev,TRUE) Greater than 73 0.5987 =1-NORM.DIST(73,Mean,Stdev,TRUE) Between 75 and 85 0.3944 =NORM.DIST(85,Mean,Stdev,TRUE)-NORM.DIST(75,Mean,Stdev,TRUE)

Normal probability calculations

Examples with standard normal

Probability calculations

Percentiles

Examples with nonstandard normal

Probability calculations

Percentiles

61.841 77.027 90.046

5th 60th

=NORM.INV(0.97,Mean,Stdev) =NORM.INV(0.6,Mean,Stdev) =NORM.INV(0.05,Mean,Stdev)

97th

= $B$18 = $B$19

75 8

Figure 5.9 Normal Calculations with Excel Functions

Figure 5.10 Typical Normal Probabilities

5-4 The Normal Distribution 2 0 7

09953_ch05_ptg01_183-241.indd 207 04/03/19 12:25 PM

can be found with NORM.INV or NORM.S.INV, using probability argument 0.975, because there must be a total probabil- ity of 0.975 to the left of x.

For example, if the relevant distribution is the standard normal, the required value of x is 1.96, found with the function NORM.S.INV(0.975). Similarly, if you want probability 0.90 in the middle and probability 0.05 in each tail, the required x is 1.645, found with the function NORM.S.INV(0.95). Remember these two numbers, 1.96 and 1.645. They occur frequently in statistical applications.

2 0 8 C H A P T E R 5 P r o b a b i l i t y a n d P r o b a b i l i t y D i s t r i b u t i o n s

5-4f Empirical Rules Revisited We introduced three empirical rules in Chapter 2 that apply to many data sets. Namely, about 68% of the data fall within one standard deviation of the mean, about 95% fall within two standard deviations of the mean, and almost all fall within three standard deviations of the mean. For these rules to hold with real data, the distribution of the data must be at least approximately symmetric and bell-shaped. Let’s look at these rules more closely.

Suppose X is normally distributed with mean m and standard deviation s. To per- form a probability calculation on X, it is useful to first standardize X and then perform the calculation on the standardized variable Z. Specifically, we will find the probability that X is within k standard deviations of its mean for k 5 1, k 5 2, and k 5 3. In general, this probability is P1m 2 ks 6 X 6 m 1 ks2. But by standardizing the values m 2 ks and m 1 ks, we obtain the equivalent probability P12k 6 Z 6 k2, where Z has a N10, 12 dis- tribution. This latter probability can be calculated in Excel with the formula

5NORM.S.DIST(k,TRUE)−NORM.S.DIST(−k,TRUE)

By substituting the values 1, 2, and 3 for k, we find the following probabilities:

P(21 < Z * 1) 5 0.6827

P(22 < Z * 2) 5 0.9545

P(23 < Z * 3) 5 0.9973

As you can see, there is virtually no chance of being beyond three standard deviations from the mean, the chances are about 19 out of 20 of being within two standard deviations of the mean, and the chances are about 2 out of 3 of being within one standard deviation of the mean. These probabilities are the basis for the empirical rules in Chapter 2. These rules more closely approximate reality as the histograms of observed data become more sym- metric and bell-shaped.

5-4g Weighted Sums of Normal Random Variables One very attractive property of the normal distribution is that if you create a weighted sum of normally distributed random variables, the weighted sum is also normally distributed. In fact, this is true even if the random variables are not independent, but we will examine only the independent case here.

Specifically, if X1 through Xn are n independent and normally distributed ran- dom variables with common mean m and common standard deviation s, then the sum X1 1 g 1 Xn is normally distributed with mean nm (sum of the means), variance ns2 (sum of the variances), and standard deviation !ns (square root of the variance). More generally, if a1 through an are any constants and each X has its own mean and standard deviation, then the weighted sum a1X1 1 g 1 an Xn is normally distributed with mean a1m1 1 g 1 an mn and variance a1

2s1 2 1 g 1 a2

n s2 n. You will need this fact to solve a

few of the problems in this chapter. (Actually, this is the correct mean even if the X’s are not independent, but in that case, the variance of the weighted sum also includes covari- ance terms and is not shown here.)

The normal distribution is the basis for the empir- ical rules introduced in Chapter 2.

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5-4 The Normal Distribution 2 0 9

5-4h Normal Distribution Examples In this section we apply the normal distribution to several business problems.

EXAMPLE

5.5 PERSONNEL TESTING AT ZTEL The personnel department of ZTel, a large communications company, is reconsidering its hiring policy. Each applicant for a job must take a standard exam, and the hire or no-hire decision depends at least in part on the result of the exam. The scores of all applicants have been examined closely. They are approximately normally distributed with mean 525 and standard deviation 55.

The current hiring policy occurs in two phases. The first phase separates all applicants into three categories: automatic accepts, automatic rejects, and maybes. The automatic accepts are those whose test scores are 600 or above. The automatic rejects are those whose test scores are 425 or below. All other applicants (the maybes) are passed on to a second phase where their previous job experience, special talents, and other factors are used as hiring criteria. The personnel manager at ZTel wants to calculate the percentage of applicants who are automatic accepts or rejects, given the current standards. She also wants to know how to change the standards to automatically reject 10% of all applicants and automatically accept 15% of all applicants.

Objective To determine test scores that can be used to accept or reject job applicants at ZTel.

Solution Let X be the test score of a typical applicant. Then historical data suggest that the distribution of X is N1525, 552. A probability such as P1X # 4252 can be interpreted as the probability that a typical applicant is an automatic reject, or it can be interpreted as the percentage of all applicants who are automatic rejects. Given this observation, the solution to ZTel’s problem appears in Figure 5.11. (See the file Personnel Decisions Finished.xlsx.) The probability that a typical applicant is automatically accepted is 0.0863, found in cell B10 with the formula

51–NORM.DIST(B7,Mean,Stdev,TRUE)

Similarly, the probability that a typical applicant is automatically rejected is 0.0345, found in cell B11 with the formula

Figure 5.11 Calculations for Personnel Example

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18

A B C D E F Personnel Decisions

Range names used: Mean of test scores Mean =$B$3 Stdev of test scores Stdev =$B$4

Current Policy Automatic accept point Automatic reject point

Percent accepted Percent rejected

=1-NORM.DIST(B7,Mean,Stdev,TRUE) =NORM.DIST(B8,Mean,Stdev,TRUE)

New Policy Percent accepted Percent rejected

Automatic accept point 582 =NORM.INV(1-B14,Mean,Stdev) Automatic reject point 455 =NORM.INV(B15,Mean,Stdev)

525 55

600 425

8.63% 3.45%

15% 10%

09953_ch05_ptg01_183-241.indd 209 04/03/19 12:25 PM

EXAMPLE

5.6 QUALITY CONTROL AT PAPERSTOCK PaperStock Company runs a manufacturing facility that produces a paper product. The fiber content of this prod- uct is supposed to be 20 pounds per 1000 square feet. (This is typical for the type of paper used in grocery bags, for example.) Because of random variations in the inputs to the process, however, the fiber content of a typical 1000-square-foot roll varies according to a N1m, s2 distribution. The mean fiber content 1m2 can be controlled—that is, it can be set to any desired level by adjusting an instrument on the machine. The variability in fiber content, as measured by the standard deviation s, is 0.10 pound when the process is “good,” but it sometimes increases to 0.15 pound when the machine goes “bad.” A given roll of this product must be rejected if its actual fiber content is less than 19.8 pounds or greater than 20.3 pounds. Calculate the probability that a given roll is rejected, for a setting of m 5 20, when the machine is “good” and when it is “bad.”

Objective To determine the machine settings that result in paper of acceptable quality at PaperStock Company.

Solution Let X be the fiber content of a typical roll. The distribution of X will be either N120, 0.102 or N120, 0.152, depending on the status of the machine. In either case, the probability that the roll must be rejected can be calculated as shown in Figure 5.12. (See the file Paper Machine Settings Finished.xlsx.) The formula for rejection in the “good” case appears in cell B12 and is spelled out to its right.

This is the sum of two probabilities: the probability of being to the left of the lower limit and the probability of being to the right of the upper limit. (See Figure 5.13.) A similar formula for the “bad” case appears in cell B13, using Stdev_bad in place of Stdev_good.

You can see that the probability of a rejected roll in the “good” case is 0.024; in the “bad” case it is 0.114. That is, when the standard deviation increases by 50% from 0.10 to 0.15, the percentage of rolls rejected more than quadruples, from 2.4% to 11.4%.

It is certainly possible that the true process mean and “good” standard deviation will not always be equal to the values in cells B3 and B4. Therefore, it is useful to see how sensitive the rejection probability is to these two parameters. You can do this with a two- way data table, as shown in Figure 5.12. The tabulated values show that the probability of rejection varies greatly even for small changes in the key inputs. In particular, a combi- nation of a badly centered mean and a large standard deviation can make the probability of rejection quite large.

To form this data table, enter the formula 5 B12 in cell B17, highlight the range B17:H25, and create a data table with row input cell B4 and column input cell B3.

2 1 0 C H A P T E R 5 P r o b a b i l i t y a n d P r o b a b i l i t y D i s t r i b u t i o n s

5NORM.DIST(B8,Mean,Stdev,TRUE)

Therefore, ZTel automatically accepts about 8.6% and rejects about 3.5% of all applicants under the current policy. To find new cutoff values that reject 10% and accept 15% of the applicants, we need the 10th and 85th percentiles of the

N1525, 552 distribution. These are 455 and 582 (rounded to the nearest integer), respectively, found in cells B17 and B18 with the formulas

5NORM.INV(1–B14,Mean,Stdev)

and

5NORM.INV(B15,Mean,Stdev)

To accomplish its objective, ZTel needs to raise the automatic rejection point from 425 to 455 and lower the automatic acceptance point from 600 to 582.

09953_ch05_ptg01_183-241.indd 210 04/03/19 12:25 PM

Figure 5.12 Calculations for Paper Quality Example

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

A B C D E F G H I J K Paper Machine Settings Range names used:

Mean =$B$3 $B$5=

Stdev_good Mean Stdev in good case

Stdev_bad =$B$4

Stdev in bad case

Reject region Lower limit Upper limit

Probability of reject in good case 0.024 =NORM.DIST(B8,Mean,Stdev_good,TRUE)+(1-NORM.DIST(B9,Mean,Stdev_good,TRUE)) in bad case 0.114 =NORM.DIST(B8,Mean,Stdev_bad,TRUE)+(1-NORM.DIST(B9,Mean,Stdev_bad,TRUE))

Data table of rejection probability as a function of the mean and good standard deviation Standard deviation

0.024 0.1 0.11 0.12 0.13 0.14 0.15 19.7 0.841 0.818 0.798 0.779 0.762 0.748 19.8 0.500 0.500 0.500 0.500 0.500 0.500 19.9 0.159 0.182 0.203 0.222 0.240 0.256

Mean 20 0.024 0.038 0.054 0.072 0.093 0.114 20.1 0.024 0.038 0.054 0.072 0.093 0.114 20.2 0.159 0.182 0.203 0.222 0.240 0.256 20.3 0.500 0.500 0.500 0.500 0.500 0.500 20.4 0.841 0.818 0.798 0.779 0.762 0.748

The point of this example is that the probability of a reject, i.e., falling outside the allowed limits, can vary greatly, not only when the mean is off target, but also when the standard deviation increases. The illustrates the famous line that “variability is the enemy.”

20 0.1

0.15

19.8 20.3

Figure 5.13 Rejection Regions for Paper Quality Example

5-4 The Normal Distribution 2 1 1

EXAMPLE

5.7 ANALYZING AN INVESTOR’S AFTER-TAX PROFIT Howard Davis invests $10,000 in a certain stock on January 1. By examining past movements of this stock and consulting with his broker, Howard estimates that the annual return from this stock, X, is normally distributed with mean 5% and stan- dard deviation 14%. Here X (when expressed as a decimal) is the profit Howard receives per dollar invested. It means that on December 31, his $10,000 will have grown to 10,000 11 1 X2 dollars. Because Howard is in the 33% tax bracket, he will then

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have to pay the Internal Revenue Service 33% of his profit, if he makes a profit. However, he doesn’t have to pay any tax if he loses money. Calculate the probability that Howard will have to pay the IRS at least $400, and calculate the probability that he won’t have to pay any tax. Also, calculate the dollar amount such that Howard’s after-tax profit is 90% certain to be less than this amount; that is, calculate the 90th percentile of his after-tax profit.

Objective To determine the after-tax profit Howard Davis can be 90% certain of earning.

Solution Howard’s before-tax profit is 10,000X dollars, so the amount he pays the IRS is 0.33110,000X2, or 3300X dollars. We want the probability that this is at least $400. Because 3300X 7 400 is the same as X 7 4/33, the probability of this outcome can be found as in Figure 5.14. (See the file Tax on Stock Return Finished.xlsx.) It is calculated in cell B8 with the formula spelled out to its right. As you can see, Howard has about a 30% chance of paying at least $400 in taxes.

Figure 5.14 Calculations for Taxable Returns Example

1 2 3 4 5 6 7 8 9

10 11 12

IHGFEDCBA Tax on Stock Return

Range names used: Amount $B$3=detsevni_tnuomAdetsevni

$B$4=naeMnaeM $B$5=vedtSvedtS

Tax $B$6=etar_xaTetar

Probability Probability of no tax

he pays at least $400 in taxes 0.305 0.360

=1-NORM.DIST(400/(Amount_invested*Tax_rate),Mean,Stdev,TRUE) =NORM.DIST(0,Mean,Stdev,TRUE)

90th percentile of stock return 22.94% =NORM.INV(0.9,Mean,Stdev) 90th percentile of after-tax return $1,537 =(1-Tax_rate)*Amount_invested*B10

000,01$ %5 %14 %33

The probability that he doesn’t have to pay any tax is easier. It is the probability the return on the stock is negative. This is 0.36, found in cell B9 with the formula shown to its right.

To answer the last question, note that the after-tax profit (when X is positive) is 67% of the before-tax profit, or 6700X dollars, and we want its 90th percentile. If this percentile is x, then we know that P16700X 6 x2 5 0.90, which is the same as P1X 6 x>67002 5 0.90. In words, we want the 90th percentile of the X distribution to be x/6700. From cell B11 of Figure 5.14, the 90th percentile is 22.94%, so the required value of x is $1,537.

is approximately normal with mean $75 and standard deviation $20. a. What is the probability that a randomly selected cus-

tomer spends less than $85 at this store? b. What is the probability that a randomly selected cus-

tomer spends between $65 and $85 at this store? c. What is the probability that a randomly selected cus-

tomer spends more than $45 at this store? d. Find the dollar amount such that 75% of all customers

spend no more than this amount. e. Find the dollar amount such that 80% of all customers

spend at least this amount. f. Find two dollar amounts, equidistant from the mean,

such that 90% of all customer purchases are between these values.

15. A machine used to regulate the amount of a certain chemical dispensed in the production of a particular type of cough syrup can be set so that it discharges an aver- age of m milliliters (ml) of the chemical in each bottle

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 13. The grades on the midterm examination given in a large

managerial statistics class are normally distributed with mean 75 and standard deviation 9. The instructor of this class wants to assign an A grade to the top 10% of the scores, a B grade to the next 10% of the scores, a C grade to the next 10% of the scores, a D grade to the next 10% of the scores, and an F grade to all scores below the 60th per- centile of this distribution. For each possible letter grade, find the lowest acceptable score within the established range. For example, the lowest acceptable score for an A is the score at the 90th percentile of this normal distribution.

14. Suppose it is known that the distribution of purchase amounts by customers entering a popular retail store

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5-4 The Normal Distribution 2 1 3

of cough syrup. The amount of chemical placed into each bottle of cough syrup is known to have a normal distribution with a standard deviation of 0.250 ml. If this machine discharges more than 2 ml of the chemi- cal when preparing a given bottle of this cough syrup, the bottle is considered to be unacceptable by industry standards. Determine the setting for m so that no more than 1% of the bottles of cough syrup prepared by this machine will be rejected.

16. Assume that the monthly sales for Toyota passenger cars follow a normal distribution with mean 5000 cars and standard deviation 1400 cars. a. There is a 1% chance that Toyota will sell more than

what number of passenger cars during the next year? (You can assume that sales in different months are probabilistically independent.)

b. What is the probability that Toyota will sell between 55,000 and 65,000 passenger cars during the next year?

17. An investor has invested in nine different investments. The dollar returns on the different investments are probabilistically independent, and each return follows a normal distribution with mean $50,000 and standard deviation $10,000. a. There is a 1% chance that the total return on the nine

investments is less than what value? b. What is the probability that the investor’s total return

is between $400,000 and $520,000? 18. Suppose that the weight of a typical American male

follows a normal distribution with m 5 180 lb and s 5 30 lb. Also, suppose 91.92% of all American males weigh more than I weigh. a. What fraction of American males weigh more than

225 pounds? b. How much do I weigh? c. If I weighed 20 pounds more than I do, what percen-

tile would I be in? 19. Assume that the length of a typical televised base-

ball game, including all the commercial timeouts, is normally distributed with mean 2.45 hours and standard deviation 0.37 hour. Consider a televised baseball game that begins at 2:00 in the afternoon. The next regularly scheduled broadcast is at 5:00. a. What is the probability that the game will cut into the

next show, that is, go past 5:00? b. If the game is over before 4:30, another half-hour

show can be inserted into the 4:30–5:00 slot. What is the probability of this occurring?

20. The amount of a soft drink that goes into a typical 12-ounce can varies from can to can. It is normally dis- tributed with an adjustable mean m and a fixed standard deviation of 0.05 ounce. (The adjustment is made to the filling machine.) a. If regulations require that cans have at least 11.9

ounces, what is the smallest mean m that can be used so that at least 99.5% of all cans meet the regulation?

b. If the mean setting from part a is used, what is the probability that a typical can has at least 12 ounces?

21. Suppose that the demands for a company’s product in weeks 1, 2, and 3 are each normally distributed. The means are 50, 45, and 65. The standard deviations are 10, 5, and 15. Assume that these three demands are probabilistically independent. a. Suppose that the company currently has 180 units in

stock, and it will not be receiving any more shipments from its supplier for at least three weeks. What is the probability that stock will run out during this three- week period?

b. How many units should the company currently have in stock so that it can be 98% certain of not running out during this three-week period? Again, assume that it won’t receive any more shipments during this period.

Level B 22. Matthew’s Bakery prepares peanut butter cookies for sale

every morning. It costs the bakery $0.50 to bake each peanut butter cookie, and each cookie is sold for $1.25. At the end of the day, leftover cookies are discounted and sold the following day at $0.40 per cookie. The daily demand (in dozens) for peanut butter cookies at this bak- ery is known to be normally distributed with mean 200 and standard deviation 60. The manager of Matthew’s Bakery is trying to determine how many dozen peanut butter cookies to make each morning to maximize the product’s contribution to bakery profits. Use simulation to find a very good, if not optimal, production plan.

23. The manufacturer of a particular bicycle model has the following costs associated with the management of this product’s inventory. In particular, the company currently maintains an inventory of 1000 units of this bicycle model at the beginning of each year. If X units are demanded each year and X is less than 1000, the excess supply, 1000 2 X units, must be stored until next year at a cost of $50 per unit. If X is greater than 1000 units, the excess demand, X 2 1000 units, must be produced separately at an extra cost of $80 per unit. Assume that the annual demand (X) for this bicycle model is normally distributed with mean 1000 and standard deviation 75. a. Find the expected annual cost associated with man-

aging potential shortages or surpluses of this product. (Hint: Use simulation to approximate the answer. An exact solution using probability arguments is beyond the level of this book.)

b. Find two annual total cost levels, equidistant from the expected value found in part a, such that 95% of all costs associated with managing potential shortages or surpluses of this product are between these values. (Continue to use simulation.)

c. Comment on this manufacturer’s annual production policy for this bicycle model in light of your findings in part b.

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2 1 4 C H A P T E R 5 P r o b a b i l i t y a n d P r o b a b i l i t y D i s t r i b u t i o n s

5-5 The Binomial Distribution The normal distribution is undoubtedly the most important probability distribution in statis- tics. Not far behind, however, is the binomial distribution. The binomial distribution is a discrete distribution that can occur in two situations: (1) when sampling from a population with only two types of members (males and females, for example), and (2) when perform- ing a sequence of identical experiments, each of which has only two possible outcomes.

24. It is widely known that many drivers on interstate highways in the United States do not observe the posted speed limit. Assume that the actual rates of speed driven by U.S. motor- ists are normally distributed with mean m mph and standard deviation 5 mph. Given this information, answer each of the following independent questions. (Hint: Use Goal Seek in parts a and b, and use the Solver add-in with no objective in part c. Solver is usually used to optimize, but it can also be used to solve equations with multiple unknowns.) a. If 40% of all U.S. drivers are observed traveling at 65

mph or more, what is the mean m? b. If 25% of all U.S. drivers are observed traveling at 50

mph or less, what is the mean m? c. Suppose now that the mean m and standard devia-

tion s of this distribution are both unknown. Further- more, it is observed that 40% of all U.S. drivers travel at less than 55 mph and 10% of all U.S. drivers travel at more than 70 mph. What must m and s be?

25. The lifetime of a certain manufacturer’s washing machine is normally distributed with mean 4 years. Only 15% of all these washing machines last at least 5 years. What is the standard deviation of the lifetime of a washing machine made by this manufacturer?

26. A fast-food restaurant sells hamburgers and chicken sand- wiches. On a typical weekday the demand for hamburgers is normally distributed with mean 313 and standard devi- ation 57; the demand for chicken sandwiches is normally distributed with mean 93 and standard deviation 22. a. How many hamburgers must the restaurant stock to be

98% sure of not running out on a given day? b. Answer part a for chicken sandwiches.

c. If the restaurant stocks 400 hamburgers and 150 chicken sandwiches for a given day, what is the prob- ability that it will run out of hamburgers or chicken sandwiches (or both) that day? Assume that the demand for hamburgers and the demand for chicken sandwiches are probabilistically independent.

d. Why is the independence assumption in part c prob- ably not realistic? Using a more realistic assumption, do you think the probability requested in part c would increase or decrease?

27. Referring to the box plots introduced in Chapter 2, the sides of the “box” are at the first and third quartiles, and the difference between these (the length of the box) is called the interquartile range (IQR). Some implementa- tions of box plots define a mild outlier as an observa- tion that is between 1.5 and 3 IQRs from the box, and an extreme outlier as an observation that is more than 3 IQRs from the box. a. If the data are normally distributed, what percent-

age of values will be mild outliers? What percent- age will be extreme outliers? Why don’t the answers depend on the mean and/or standard deviation of the distribution?

b. Check your answers in part a with simulation. Simu- late a large number of normal random numbers (you can choose any mean and standard deviation), and count the number of mild and extreme outliers with appropriate formulas. Do these match, at least approx- imately, your answers to part a?

What the Binomial Distribution Describes

Unlike the normal distribution, which can describe many types of random phenomena, the binomial distribution is relevant for a very common and specific situation: the number of successes in a fixed number of trials, where the trials are probabilistically independent and the probability of success remains constant across trials. Whenever this situation occurs, the binomial distribution is the relevant distribution for the number of successes.

Fundamental Insight

For the latter case, imagine any experiment that can be repeated many times under identical conditions. It is common to refer to each repetition of the experiment as a trial.

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5-5 The Binomial Distribution 2 1 5

We assume that the outcomes of successive trials are probabilistically independent of one another and that each trial has only two possible outcomes. We label these two possibil- ities generically as success and failure. In any particular application the outcomes might be Democrat/Republican, defective/nondefective, went bankrupt/remained solvent, and so on. We label the probability of a success on each trial as p, and the probability of a failure as 1 2 p. We let n be the number of trials.

Consider a situation where there are n independent, identical trials, where the probability of a success on each trial is p and the probability of a failure is 1 2 p. Define X to be the random number of successes in the n trials. Then X has a bino- mial distribution with parameters n and p.

For example, the binomial distribution with parameters 100 and 0.3 is the distribution of the number of successes in 100 trials when the probability of success is 0.3 on each trial. A simple example that you can keep in mind throughout this section is the number of heads you would see if you flipped a coin n times. Assuming the coin is well balanced, the relevant distribution is binomial with parameters n and p 5 0.5. This coin-flipping example is often used to illustrate the binomial distribution because of its simplicity, but you will see that the binomial distribution also applies to many important business situations.

To understand how the binomial distribution works, consider the coin-flipping exam- ple with n 5 3. If X represents the number of heads in three flips of the coin, then the possible values of X are 0, 1, 2, and 3. You can find the probabilities of these values by con- sidering the eight possible outcomes of the three flips: (T,T,T), (T,T,H), (T,H,T), (H,T,T), (T,H,H), (H,T,H), (H,H,T), and (H,H,H). Because of symmetry (the well-balanced prop- erty of the coin), each of these eight possible outcomes must have the same probability, so each must have probability 1>8. Next, note that one of the outcomes has X 5 0, three outcomes have X 5 1, three outcomes have X 5 2, and one outcome has X 5 3. There- fore, the probability distribution of X is

P1X 5 02 5 1>8, P1X 5 12 5 3>8, P1X 5 22 5 3>8, P1X 5 32 5 1>8

This is a special case of the binomial distribution, with n 5 3 and p 5 0.5. In general, where n can be any positive integer and p can be any probability between 0 and 1, there is a rather complex formula for calculating P 1X 5 k2 for any integer k from 0 to n. Instead of presenting this formula, we will discuss how to calculate binomial probabilities in Excel. You do this with the BINOM.DIST function. The general form of this function is

5BINOM.DIST(k, n, p,cum)

The middle two arguments are the number of trials n and the probability of success p on each trial. The first parameter k is an integer number of successes that you specify. The last parameter, cum, is either TRUE or FALSE. It is TRUE (or 1) if you want the probability of less than or equal to k successes, and it is FALSE (or 0) if you want the probability of exactly k successes. We illustrate typical binomial calculations, including those involving the BINOM.INV function for percentile calculations, in Example 5.8.

Old Binomial Functions

Prior to Excel 2010, the binomial functions were BINOMDIST and CRITBI- NOM. They are exactly the same as the new BINOM.DIST and BINOM.INV functions. Microsoft kept these old functions for backward compatibility, so you can continue to use them if you prefer.

Excel Tip

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With this in mind, the probabilities requested in (a) through (f) become:

a. P1X 5 582 b. P1X # 652 c. P1X 6 702 5 P1X # 692 d. P1X $ 592 5 1 2 P1X 6 592 5 1 2 P1X # 582 e. P1X 7 652 5 1 2 P1X # 652 f. P155 # X # 652 5 P1X # 652 2 P1X 6 552 5 P1X # 652 2 P1X # 542

Note how we have manipulated each of these so that it includes only terms of the form P1X 5 k2 or P1X # k2 for appropriate values of k. These are the types of probabilities that can be handled directly by the BINOM.DIST function. The answers appear in the range B7:B12, and the corresponding formulas are shown (as labels) to their right.

The probabilities requested in (g) through (i) involve failures rather than successes. But because each trial results in either a success or a failure, the number of failures is also binomially distributed, with parameters n and 1 2 p 5 0.4. So in rows 14

EXAMPLE

5.8 BINOMIAL CALCULATIONS IN EXCEL Suppose that 100 identical batteries are inserted in identical flashlights. Each flashlight takes a single battery. After eight hours of continuous use, a given battery is still operating with probability 0.6 or has failed with probability 0.4. Let X be the number of successes in these 100 trials, where a success means that the battery is still functioning. Find the probabilities of the following events: (a) exactly 58 successes, (b) no more than 65 successes, (c) less than 70 successes, (d) at least 59 successes, (e) greater than 65 successes, (f) between 55 and 65 successes (inclusive), (g) exactly 40 failures, (h) at least 35 failures, and (i) less than 42 failures. Then find the 95th percentile of the distribution of X.

Objective To use Excel’s BINOM.DIST and BINOM.INV functions for calculating binomial probabilities and percentiles.

Solution Figure 5.15 shows the solution to all of these problems. (See the file Binomial Calculations Finished.xlsx.) The probabilities requested in parts (a) through (f) all involve the number of successes X. The key to these is the wording of phrases such as “no more than,” “greater than,” and so on. In particular, you have to be careful to distinguish between probabilities such as P1X 6 k2 and P1X # k2. The latter includes the possibility of having X 5 k and the former does not.

Figure 5.15 Typical Binomial Calculations 1

HGFEDCBA Binomial Probability Calculations

Range names used: n =$B$3 p =$B$4

=BINOM.DIST(58,n,p,FALSE) =BINOM.DIST(65,n,p,TRUE) =BINOM.DIST(69,n,p,TRUE) =1-BINOM.DIST(58,n,p,TRUE) =1-BINOM.DIST(65,n,p,TRUE) =BINOM.DIST(65,n,p,TRUE)-BINOM.DIST(54,n,p,TRUE)

=BINOM.DIST(40,n,1-p,FALSE) =1-BINOM.DIST(34,n,1-p,TRUE) =BINOM.DIST(41,n,1-p,TRUE)

=BINOM.DIST(A20,n,p,TRUE) (Copy down)

=BINOM.INV(n,p,B27)

Number of Probability of success on each trial

ytilibaborPtnevE Exactly 58 2470.0sesseccus No more than 65 7968.0sesseccus Less than 70 2579.0sesseccus At least 59 5226.0sesseccus Greater than 65 3031.0sesseccus Between 55 and 65 successes (inclusive) 0.7386

Exactly 40 2180.0seruliaf At least 35 7968.0seruliaf Less than 42 5226.0seruliaf

Finding the 95th percentile (trial and error) Trial values CumProb

65 0.8697 66 0.9087 67 0.9385 68 0.9602 69 0.9752 70 0.9852

68 0.95

001slairt 0.6

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through 16 the requested probabilities are calculated in exactly the same way, except that 1-p is substituted for p in the third argument of the BINOM.DIST function.

Finally, to calculate the 95th percentile of the distribution of X, you can proceed by trial and error. For each value k from 65 to 70, the probability P1X # k2 is calculated in column B with the BINOM.DIST function. As you can see, there is no value k such that P1X # k2 5 0.95 exactly. Specifically, P1X # 672 is slightly less than 0.95 and P1X # 682 is slightly greater than 0.95. Therefore, the meaning of the “95th percentile” is somewhat ambiguous. If you want the largest value k such that P1X # k2 # 0.95, then this k is 67. If instead you want the smallest value k such that P1X # k2 $ 0.95, then this value is 68. The latter interpretation is the one usually accepted for binomial percentiles.

In fact, Excel has another built-in function, BINOM.INV for finding this value of k. This function is illustrated in row 27 of Figure 5.15 in cell A27. This BINOM.INV function returns 68, the smallest value k such that P1X # k2 $ 0.95 for this binomial distribution.

5-5 The Binomial Distribution 2 1 7

5-5a Mean and Standard Deviation of the Binomial Distribution

It can be shown that the mean and standard deviation of a binomial distribution with parameters n and p are given by the following equations.

Mean and Standard Deviation of the Binomial Distribution

E1X2 5 np (5.14)

Stdev1X2 5 !np11 2 p2 (5.15)

The formula for the mean is quite intuitive. For example, if you observe 100 trials, each with probability of success 0.6, your best guess for the number of successes is 10010.62 5 60. The standard deviation is less obvious but still very useful. It indicates how far the actual number of successes is likely to deviate from the mean. In this case the standard deviation is !10010.62 10.42 5 4.90.

Fortunately, the empirical rules discussed in Chapter 2 also apply, at least approxi- mately, to the binomial distribution. That is, there is about a 95% chance that the actual number of successes will be within two standard deviations of the mean, and there is almost no chance that the number of successes will be more than three standard deviations from the mean. So for this example, it is very likely that the number of successes will be in the range of approximately 50 to 70, and it is very unlikely that there will be fewer than 45 or more than 75 successes.

This reasoning is extremely useful. It provides a rough estimate of the number of successes you are likely to observe. Suppose 1000 parts are sampled randomly from an assembly line and, based on historical performance, the percentage of parts with some type of defect is about 5%. Translated into a binomial model, each of the 1000 parts, independently of the others, has some type of defect with probability 0.05. Would it be surprising to see, say, 75 parts with a defect? The mean is 100010.052 5 50 and the stan- dard deviation is !100010.052 10.952 5 6.89. Therefore, the number of parts with defects is approximately 95% certain to be within 50 { 216.892 , or approximately from 36 to 64. Because 75 is slightly beyond three standard deviations from the mean, it is highly unlikely that there would be 75 (or more) defective parts.

5-5b The Binomial Distribution in the Context of Sampling The binomial distribution also applies to sampling from a population with two types of members. Let’s say these two types are men and women, although in applications they might be Democrats and Republicans, users of our product and nonusers, and so on. We assume that the population has N members, of whom NM are men and NW are women (where NM 1 NW 5 N). If you sample n of these randomly, you are typically interested in

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the composition of the sample. You might expect the number of men in the sample to be binomially distributed with parameters n and p 5 NM>N, the fraction of men in the popu- lation. However, this depends on how the sampling is performed.

If sampling is done without replacement, each member of the population can be sam- pled only once. That is, once a person is sampled, his or her name is struck from the list and cannot be sampled again. If sampling is done with replacement, then it is possible, although maybe not likely, to select a given member of the population more than once. Most real-world sampling is performed without replacement. There is no point in obtaining infor- mation from the same person more than once. However, the binomial model applies only to sampling with replacement. Because the composition of the remaining population keeps changing as the sampling progresses, the binomial model provides only an approximation if sampling is done without replacement. If there is no replacement, the value of p, the pro- portion of men in this case, does not stay constant, a requirement of the binomial model. The appropriate distribution for sampling without replacement is called the hypergeometric distribution, a distribution we will not discuss here.5

If n is small relative to N, however, the binomial distribution is a very good approxi- mation to the hypergeometric distribution and can be used even if sampling is performed without replacement. A rule of thumb is that if n is no greater than 10% of N, that is, no more than 10% of the population is sampled, then the binomial model can be used safely. Of course, most national polls sample considerably less than 10% of the population. In fact, they often sample only about a thousand people from the hundreds of millions in the entire population. The bottom line is that in most real-world sampling contexts, the bino- mial model is perfectly adequate.

5-5c The Normal Approximation to the Binomial If you graph the binomial probabilities, you will see an interesting phenomenon—namely, the graph begins to look symmetric and bell-shaped when n is fairly large and p is not too close to 0 or 1. An example is illustrated in Figure 5.16 with the parameters n 5 30 and p 5 0.4. Generally, if np 7 5 and n11 2 p2 7 5, the binomial distribution can be approx- imated well by a normal distribution with mean np and standard deviation !np11 2 p2. Before Excel functions were available, this fact was used to find approximate binomial probabilities from normal tables. Although this is no longer necessary, it is still useful to know that a binomial distribution resembles a normal distribution for large n.

If n is large and p is not too close to 0 or l, the binomial distribution is bell-shaped and can be approxi- mated well by the normal distribution.

Continuity Correction

Because the normal distribution is continuous and the binomial distribution is discrete, the normal approximation to the binomial can be improved slight- ly with a continuity correction. If you want to approximate a binomial prob- ability such as P136 # X # 452, expand the interval by 0.5 on each end in the normal approximation. That is, approximate with the normal probability P135.5 # X # 45.52. Similarly, approximate binomial P1X # 452 with normal P1X # 45.52, or binomial P1X $ 362 with normal P1X $ 35.52. Admittedly, this continuity correction is mostly of historical interest. With Excel’s binomial functions, there is no need to resort to a normal approximation.

Excel Tip

One practical consequence of the normal approximation to the binomial is that the empirical rules can be applied. That is, when the binomial distribution is approximately symmetric and bell-shaped, there is about a 68% chance that the number of successes will

5 Excel has a function HYPGEOM.DIST for sampling without replacement that works much like the BINOM. DIST function. You can look it up under the Statistical category of Excel functions.

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5-5 The Binomial Distribution 2 1 9

be within one standard deviation of the mean. Similarly, there is about a 95% chance that the number of successes will be within two standard deviations of the mean, and the num- ber of successes will almost surely be within three standard deviations of the mean. Here, the mean is np and the standard deviation is !np11 2 p2.

5-5d Binomial Distribution Examples The binomial distribution finds many applications in the business world and elsewhere. We discuss a few typical applications in this section.

Figure 5.16 Bell-shaped Binomial Distribution

Relationship Between Normal and Binomial Distributions

If you look at a graph of a binomial distribution when n is fairly large and p is not too close to 0 or 1, you will see that the distribution is bell-shaped. This is no accident. It can be proven mathematically that the normal distribution provides a very good approximation to the binomial under these conditions (n large, p not too close to 0 or 1). One implication is that the empirical rules from Chapter 2 apply very well to binomial distributions, using the mean and standard deviation in Equations (5.14) and (5.15). For example, there is about a 95% chance that the number of successes will be within two standard deviations of the mean.

Fundamental Insight

EXAMPLE

5.9 IS THIS MUTUAL FUND REALLY A WINNER? An investment broker at the Michaels & Dodson Company claims that he has found a real winner. He has tracked a mutual fund that has beaten a standard market index in 37 of the past 52 weeks. Could this be due to chance, or has he really found a winner?

Objective To determine the probability of a mutual fund outperforming a standard market index at least 37 out of 52 weeks.

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Solution The broker is no doubt tracking a lot of mutual funds, and he is probably reporting only the best of these. Therefore, we will check whether the best of many mutual funds could do at least this well purely by chance. To do this, we first specify what we mean by “purely by chance.” This means that each week, a given fund has a fifty-fifty chance of beating the market index, independently of performance in other weeks. In other words, the number of weeks where a given fund outperforms the market index is binomially distributed with n 5 52 and p 5 0.5. With this in mind, cell B6 of Figure 5.17 shows the probability that a given fund does at least as well—beats the market index at least 37 out of 52 weeks—as the reported fund. (See the Beating the Market Finished.xlsx file.) Because P1X $ 372 5 1 2 P1X # 362, the relevant formula is

51–BINOM.DIST(B3−1,B4,0.5,TRUE)

Obviously, this probability, 0.00159, is quite small. A single fund isn’t likely to beat the market this often purely by chance.

Figure 5.17 Binomial Calculations for Investment Example

1 2 3 4 5

6 7 8

9 10 11 12 13 14 15 16 17 18

A B C D E F G Beating the market

Weeks beating market index Total number of weeks

Probability of doing at least this well by chance 0.00159 =1-BINOM.DIST(B3-1,B4,0.5,TRUE)

Number of mutual funds Probability of at least one doing at least this well 0.471 =1-BINOM.DIST(0,B8,B6,TRUE)

Two-way data table of the probability in B9 as a function of values in B3 and B8 Number of weeks beating the market index

0.471 36 37 38 39 40 Number of mutual funds 200 0.542 0.273 0.113 0.040 0.013

300 0.690 0.380 0.164 0.060 0.019 400 0.790 0.471 0.213 0.079 0.025 500 0.858 0.549 0.258 0.097 0.031 600 0.904 0.616 0.301 0.116 0.038

37 52

400

However, the probability that the best of many mutual funds does at least this well is much larger. To calculate this prob- ability, assume that 400 funds are being tracked, and let Y be the number of these that beat the market at least 37 of 52 weeks. Then Y is also binomially distributed, with parameters n 5 400 and p 5 0.00159, the probability calculated previously. To see whether any of the 400 funds beats the market at least 37 of 52 weeks, calculate P1Y $ 12 5 1 2 P1Y 5 02 in cell B9 with the formula

51–BINOM.DIST(0,B8,B7,TRUE)

(Can you see why the fourth argument could be TRUE or FALSE?) The resulting probability is nearly 0.5—that is, there is nearly a fifty-fifty chance that at least one of 400 funds will do as well as the reported fund. This certainly casts doubt on the broker’s claim that he has found a real winner. It is more likely that his star fund just got lucky and will perform no better than average in succeeding weeks.

To see how the probability in cell B9 depends on the level of success of the reported fund (the value in cell B3) and the number of mutual funds being tracked (in cell B8), a two-way data table has been created in the range B13:G18. (The formula in cell B13 is =B9, the row input cell is B3, and the column input cell is B8.) As you saw, beating the market 37 times out of 52 is no big deal with 400 funds, but beating it 40 times out of 52, even with 600 funds, is something worth reporting. The probability of this happening purely by chance is only 0.038, or less than 1 out of 25.

2 2 0 C H A P T E R 5 P r o b a b i l i t y a n d P r o b a b i l i t y D i s t r i b u t i o n s

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5-5 The Binomial Distribution 2 2 1

The next example requires a normal calculation to find a probability p, which is then used in a binomial calculation.

EXAMPLE

5.10 DAILY SALES AT A SUPERMARKET Customers at a supermarket spend varying amounts. Historical data indicate that the amount spent per customer is normally distributed with mean $85 and standard deviation $30. If 500 customers shop in a given day, calculate the mean and standard deviation of the number who spend at least $100. Then calculate the probability that at least 30% of all customers spend at least $100.

Objective To use the normal and binomial distributions to calculate the typical number of customers who spend at least $100 per day and the probability that at least 30% of all 500 daily customers spend at least $100.

Solution Both questions involve the number of customers who spend at least $100. Because the amounts spent are normally distributed, the probability that a typical customer spends at least $100 is found with the NORM.DIST function. This probability, 0.309, appears in cell B7 of Figure 5.18. (See the file Supermarket Spending Finished.xlsx.) It is calculated with the formula

51-NORM.DIST(100,B4,B5,TRUE)

This probability is then used as the parameter p in a binomial model. The mean and standard deviation of the number who spend at least $100 are calculated in cells B13 and B14 as np and !np11 2 p2 using n 5 500, the number of shoppers, and p 5 0.309. The expected number who spend at least $100 is slightly greater than 154, and the standard deviation of this num- ber is slightly greater than 10.

To answer the second question, note that 30% of 500 customers is 150 customers. Then the probability that at least 30% of the customers spend at least $100 is the probability that a binomially distributed random variable, with n 5 500 and p 5 0.309, is at least 150. This binomial probability, which turns out to be about 2>3, is calculated in cell B16 with the formula

51-BINOM.DIST(0.3*B10-1,B10,B7,TRUE)

Note that the first argument evaluates to 149. This is because the probability of at least 150 customers is one minus the probability of less than or equal to 149 customers.

FEDCBA

Supermarket spending

Amount spent per customer (normally distributed) Mean StDev

Probability that a customer spends at least $100 0.309 =1-NORM.DIST(100,B4,B5,TRUE)

Number of customers

Mean and stdev of number who spend at least $100 =B10*B7154.27Mean =SQRT(B10*B7*(1-B7))10.33StDev

Probability at least 30% spend at least $100 0.676 =1-BINOM.DIST(0.3*B10-1,B10,B7,TRUE)

$85 $30

005

Figure 5.18 Calculations for Supermarket Example

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EXAMPLE

5.11 OVERBOOKING BY AIRLINES This example presents a simplified version of calculations used by airlines when they overbook flights. They realize that a certain percentage of ticketed passengers will cancel at the last minute. Therefore, to avoid empty seats, they sell more tickets than there are seats, hoping that just about the right number of passengers show up. We assume that the no-show rate is 10%. In binomial terms, we assume that each ticketed passenger, independently of the others, shows up with probability 0.90 and cancels with probability 0.10.

For a flight with 200 seats, the airline wants to see how sensitive various probabilities are to the number of tickets it issues. In particular, it wants to calculate (a) the probability that more than 205 passengers show up, (b) the probability that more than 200 passengers show up, (c) the probability that at least 195 seats are filled, and (d) the probability that at least 190 seats are filled. The first two of these are “bad” events from the airline’s perspective; they mean that some customers will be bumped from the flight. The last two events are “good” in the sense that the airline wants most of the seats to be occupied.

Objective To assess the benefits and drawbacks of airline overbooking.

Solution To solve the airline’s problem, we use the BINOM.DIST function and a data table. The solution appears in Figure 5.19. (See the file Airline Overbooking Finished.xlsx.) For any number of tickets issued in cell B6, the required probabilities are calcu- lated in row 10. For example, the formulas in cells B10 and D10 are

51-BINOM.DIST(205,NTickets,1-PNoShow,TRUE)

Figure 5.19 Binomial Calculations for Overbooking Example

1 2 3 4 5 6 7 8

9 10 11 12

13 14 15 16 17 18 19 20 21 22 23 24

A B C D E Airline overbooking Range names used:

NTickets =$B$6 Number of seats PNoShow =$B$4 Probability of no-show

Number of tickets issued

Required probabilities More than 200

show up More than

205 show up At least 190

seats filled At least 195

seats filled

0.001 0.050 0.421 0.820

Data table showing sensitivity of probabilities to number of tickets issued

Number of tickets issued More than

205 show up More than 200

show up At least 195

seats filled At least 190

seats filled 0.001 0.050 0.421 0.820

0.384

0.370 0.607

206 0.000 0.000 0.012 0.171 209 0.000 0.001 0.064 212 0.000 0.009 0.201 0.628 215 0.001 0.050 0.421 0.820 218 0.013 0.166 0.659 0.931 221 0.064 224 0.194 227 0.406 0.802 0.981 0.999 230 0.639 0.920 0.995 1.000 233 0.822 0.974 0.999 1.000

0.9950.939 0.9780.839

002 0.1

215

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5-5 The Binomial Distribution 2 2 3

The following example concerns a real problem that occurs every time you watch election returns on TV. This problem is of particular interest in light of the highly unusual events that took place during election night television coverage of the U.S. presidential election in 2000, where the networks declared Al Gore an early winner in at least one state that he eventually lost. The basic question is how soon the networks can declare one of the candidates the winner, based on early voting returns. Our example is somewhat unrealistic because it ignores the possibility that early tabulations can be biased one way or the other. For example, the earliest reporting precincts might be known to be more heavily in favor of the Democrat than the population in general. Nevertheless, the example indicates, at least approximately, why the networks are able to make early conclusions based on such seemingly small amounts of data.

and

51-BINOM.DIST(194,NTickets,1-PNoShow,TRUE)

Note that the condition “more than” requires a slightly different calculation from “at least.” The probability of more than 205 is one minus the probability of less than or equal to 205, whereas the probability of at least 195 is one minus the probabil- ity of less than or equal to 194. Also, note that a passenger who shows up is called a success. Therefore, the third argument of each BINOM.DIST function is one minus the no-show probability.

To see how sensitive these probabilities are to the number of tickets issued, a one-way data table was created at the bottom of the spreadsheet. It is one-way because there is only one input, the number of tickets issued, even though four output prob- abilities are tabulated. (To create the data table, list several possible numbers of tickets issued along the side in column A and create links in row 14 to the probabilities in row 10. That is, enter the formula =B10 in cell B14 and copy it across row 14. Then form a data table using the range A14:E24, no row input cell, and column input cell B6.)

The results are as expected. As the airline issues more tickets, there is a larger chance of having to bump passengers from the flight, but there is also a larger chance of filling most seats. In reality, the airline has to make a trade-off between these two, taking its various costs and prices into account.

EXAMPLE

5.12 PROJECTING ELECTION WINNERS FROM EARLY RETURNS We assume that there are N voters in the population, of whom NR will vote for the Republican and ND will vote for the Dem- ocrat. The eventual winner will be the Republican if NR 7 ND and will be the Democrat otherwise, but we won’t know which until all of the votes are tabulated. (To simplify the example, we assume there are only two candidates and that the election will not end in a tie.) Let’s suppose that a small percentage of the votes have been counted and the Republican is currently ahead 540 to 460. On what basis can the networks declare the Republican the winner, especially if there are millions of voters in the population?

Objective To use a binomial model to determine whether early returns reflect the eventual winner of an election between two candidates.

Solution Let n 5 1000 be the total number of votes that have been tabulated. If X is the number of Republican votes so far, we are given that X 5 540. Now we pose the following question. If the Democrat were going to be the eventual winner, that is, ND 7 NR, and we randomly sampled 1000 voters from the population, how likely is it that at least 540 of these voters would be in favor of the Republican? If this is very unlikely, then the only reasonable conclusion is that the Democrat will not be the eventual winner. This is the reasoning the networks might use to declare the Republican the winner so early in the tabulation.

We use a binomial model to see how unlikely the event “at least 540 out of 1000 ” is, assuming that the Democrat will be the eventual winner. We need a value for p, the probability that a typical vote is for the Republican. This probability should be

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the proportion of voters in the entire population who favor the Republican. All we know is that this probability is less than 0.5, because we have assumed that the Democrat will eventually win. In Figure 5.20, we show how the probability of at least 540 out of 1000 varies with values of p less than, but close to, 0.5. (See the file Election Returns Finished.xlsx.)

We enter a trial value of 0.49 for p in cell B3 and then calculate the required probability in cell B9 with the formula

=1-BINOM.DIST(B6-1,B5,B3,TRUE)

Then we use this to create the data table at the bottom of the spreadsheet. This data table tabulates the probability of the given lead (at least 540 out of 1000) for various values of p less than 0.5. As shown in the last few rows, even if the eventual outcome were going to be a virtual tie—with the Democrat slightly ahead—there would still be very little chance of the Republican being at least 80 votes ahead so far. But because the Republican is currently ahead by 80 votes, the networks feel safe in declaring the Republican the winner. Admittedly, the probability model they use is more complex than our simple binomial model, but the idea is the same.

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 28. Many vehicles used in space travel are constructed with

redundant systems to protect flight crews and their valu- able equipment. In other words, backup systems are included within many vehicle components so that if one or more systems fail, backup systems will assure the safe operation of the given component and thus the entire vehicle. For example, consider one particular com- ponent of the U.S. space shuttle that has n duplicated systems (i.e., one original system and n 2 1 backup sys- tems). Each of these systems functions, independently of the others, with probability 0.98. This shuttle component functions successfully provided that at least one of the n systems functions properly.

a. Find the probability that this shuttle component func- tions successfully if n 5 2.

b. Find the probability that this shuttle component functions successfully if n 5 4.

c. What is the minimum number n of duplicated systems that must be incorporated into this shuttle component to ensure at least a 0.9999 probability of successful operation?

29. Suppose that a popular hotel for vacationers in Orlando, Florida, has a total of 300 identical rooms. As many major airline companies do, this hotel has adopted an overbooking policy in an effort to maximize the usage of its available lodging capacity. Assume that each potential hotel customer holding a room reservation, independently of other customers, cancels the reserva- tion or simply does not show up at the hotel on a given night with probability 0.15.

Figure 5.20 Binomial Calculations for Voting Example

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18

FEDCBA Election returns

Population proportion for Republican

Votes tabulated so far Votes for Republican so far

P(at least this many R votes) - binomial 0.0009 =1-BINOM.DIST(B6-1,B5,B3,TRUE)

Data table showing sensitivity of this probability to population proportion for Republican Population proportion for Republican

0.0009 0.490 0.0009 0.492 0.0013 0.494 0.0020 0.496 0.0030 0.498 0.0043 0.499 0.0052

Probability

0.49

1000 540

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5-5 The Binomial Distribution 2 2 5

a. Find the largest number of room reservations that this hotel can book and still be at least 95% sure that everyone who shows up at the hotel will have a room on a given night.

b. Given that the hotel books the number of reservations found in part a, find the probability that at least 90% of the available rooms will be occupied on a given night.

c. Given that the hotel books the number of reservations found in part a, find the probability that at most 80% of the available rooms will be occupied on a given night.

d. How does your answer to part a change as the required assurance rate increases from 95% to 97%? How does your answer to part a change as the required assurance rate increases from 95% to 99%?

e. How does your answer to part a change as the cancel- lation rate varies between 5% and 25% (in increments of 5%)? Assume now that the required assurance rate remains at 95%.

30. A production process manufactures items with weights that are normally distributed with mean 15 pounds and standard deviation 0.1 pound. An item is considered to be defective if its weight is less than 14.8 pounds or greater than 15.2 pounds. Suppose that these items are currently produced in batches of 1000 units. a. Find the probability that at most 5% of the items in a

given batch will be defective. b. Find the probability that at least 90% of the items in a

given batch will be acceptable. c. How many items would have to be produced in a

batch to guarantee that a batch consists of no more than 1% defective items?

31. Past experience indicates that 30% of all individu- als entering a certain store decide to make a purchase. Using (a) the binomial distribution and (b) the normal approximation to the binomial, find that probability that 10 or more of the 30 individuals entering the store in a given hour will decide to make a purchase. Compare the results obtained using the two different approaches. Under what conditions will the normal approximation to this binomial probability become even more accurate?

32. Suppose that the number of ounces of soda put into a soft-drink can is normally distributed with m 5 12.05 ounces and s 5 0.03 ounce. a. Legally, a can must contain at least 12 ounces of soda.

What fraction of cans will contain at least 12 ounces of soda?

b. What fraction of cans will contain less than 11.9 ounces of soda?

c. What fraction of cans will contain between 12 and 12.08 ounces of soda?

d. One percent of all cans will weigh more than what value?

e. Ten percent of all cans will weigh less than what value?

f. The soft-drink company controls the mean weight in a can by setting a timer. For what mean should the timer be set so that only 1 in 1000 cans will be underweight?

g. Every day the company produces 10,000 cans. The government inspects 10 randomly chosen cans each day. If at least two are underweight, the company is fined $10,000. Given that m 5 12.05 ounces and s 5 0.03 ounce, what is the probability that the com- pany will be fined on a given day?

33. Suppose that 53% of all registered voters prefer presi- dential candidate Smith to presidential candidate Jones. (You can substitute the names of the most recent presi- dential candidates.) a. In a random sample of 100 voters, what is the proba-

bility that the sample will indicate that Smith will win the election (that is, there will be more votes in the sample for Smith)?

b. In a random sample of 100 voters, what is the proba- bility that the sample will indicate that Jones will win the election?

c. In a random sample of 100 voters, what is the prob- ability that the sample will indicate a dead heat (fifty-fifty)?

d. In a random sample of 100 voters, what is the prob- ability that between 40 and 60 (inclusive) voters will prefer Smith?

34. Assume that, on average, 95% of all ticket holders show up for a flight. If a plane seats 200 people, how many tickets should be sold to make the chance of an over- booked flight as close as possible to 5%?

35. Suppose that 55% of all people prefer Coke to Pepsi. We randomly choose 500 people and ask them if they pre- fer Coke to Pepsi. What is the probability that our sur- vey will (erroneously) indicate that Pepsi is preferred by more people than Coke? Does this probability increase or decrease as we take larger and larger samples? Why?

36. Suppose that 4% of all tax returns are audited. In a group of n tax returns, consider the probability that at most two returns are audited. How large must n be before this probability is less than 0.01?

37. Suppose that the height of a typical American female is normally distributed with m 5 64 inches and s 5 4 inches. We observe the height of 500 American females. a. What is the probability that fewer than 35 of the 500

women will be less than 58 inches tall? b. Let X be the number of the 500 women who are less

than 58 inches tall. Find the mean and standard devia- tion of X.

Level B 38. Many firms utilize sampling plans to control the quality

of manufactured items ready for shipment. To illustrate the use of a sampling plan, suppose that a particular com- pany produces and ships electronic computer chips in lots, each lot consisting of 1000 chips. This company’s

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sampling plan specifies that quality control personnel should randomly sample 50 chips from each lot and accept the lot for shipping if the number of defective chips is four or fewer. The lot will be rejected if the num- ber of defective chips is five or more. a. Find the probability of accepting a lot as a func-

tion of the actual fraction of defective chips. In particular, let the actual fraction of defective chips in a given lot equal any of 0.02, 0.04, 0.06, 0.08, 0.10, 0.12, 0.14, 0.16, 0.18. Then compute the lot acceptance probability for each of these lot defec- tive fractions.

b. Construct a graph showing the probability of lot acceptance for each of the lot defective fractions, and interpret your graph.

c. Repeat parts a and b under a revised sampling plan that calls for accepting a given lot if the number of defective chips found in the random sample of 50 chips is five or fewer. Summarize any notable differ- ences between the two graphs.

39. A standardized test consists entirely of multiple-choice questions, each with five possible choices. You want to ensure that a student who randomly guesses on each question will obtain an expected score of zero. How can you accomplish this?

40. In the current tax year, suppose that 5% of the millions of individual tax returns are fraudulent. That is, they contain errors that were purposely made to cheat the government. a. Although these errors are often well concealed, let’s

suppose that a thorough IRS audit will uncover them. If a random 250 tax returns are audited, what is the probability that the IRS will uncover at least 15 fraud- ulent returns?

b. Answer the same question as in part a, but this time assume there is only a 90% chance that a given fraud- ulent return will be spotted as such if it is audited.

41. Suppose you work for a survey research company. In a typical survey, you mail questionnaires to 150 companies. Of course, some of these companies might decide not to respond. Assume that the nonresponse rate is 45%; that is, each company’s probability of not responding, inde- pendently of the others, is 0.45. a. If your company requires at least 90 responses for a

valid survey, find the probability that it will get this

many. Use a data table to see how your answer varies as a function of the nonresponse rate (for a reasonable range of response rates surrounding 45%).

b. Suppose your company does this survey in two “waves.” It mails the 150 questionnaires and waits a certain period for the responses. As before, assume that the nonresponse rate is 45%. However, after this initial period, your company follows up (by tele- phone, say) on the nonrespondents, asking them to please respond. Suppose that the nonresponse rate on this second wave is 70%; that is, each original non- respondent now responds with probability 0.3, inde- pendently of the others. Your company now wants to find the probability of obtaining at least 110 responses total. It turns out that this is a difficult probability to calculate directly. So instead, approximate it with simulation.

42. Suppose you are sampling from a large population, and you ask the respondents whether they believe men should be allowed to take paid paternity leave from their jobs when they have a new child. Each person you sam- ple is equally likely to be male or female. The popula- tion proportion of females who believe males should be granted paid paternity leave is 56%, and the population proportion of males who favor it is 48%. If you sam- ple 200 people and count the number who believe males should be granted paternity leave, is this number bino- mially distributed? Explain why or why not. Would your answer change if you knew your sample was going to consist of exactly 100 males and 100 females?

43. A woman claims that she is a fortune-teller. Specif- ically, she claims that she can predict the direction of the change (up or down) in the Dow Jones Industrial Average for the next 10 days. For example, one possible prediction might be U, U, D, U, D, U, U, D, D, D. (You can assume that she makes all 10 predictions right now, although that does not affect your answer to the ques- tion.) Obviously, you are skeptical, thinking that she is just guessing, so you would be surprised if her pre- dictions are accurate. Which would surprise you more: (1) she predicts at least 8 out of 10 correctly, or (2) she predicts at least 6 out of 10 correctly on each of four separate occasions? Answer by assuming that (1) she is really guessing and (2) each day the Dow is equally likely to go up or down.

5-6 The Poisson and Exponential Distributions The final two distributions in this chapter are called the Poisson and exponential distributions. In most statistical applications, including those in the rest of this book, these distributions play a much less important role than the normal and binomial distributions. For this reason, we will not analyze them in much detail. However, in many applied man- agement science models, the Poisson and exponential distributions are key distributions.

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5-6 The Poisson and Exponential Distributions 2 2 7

For example, much of the study of probabilistic inventory models, queueing models, and reliability models relies heavily on these two distributions.

5-6a The Poisson Distribution The Poisson distribution is a discrete distribution. It usually applies to the number of events occurring within a specified period of time or space. Its possible values are all of the nonnegative integers: 0, 1, 2, and so on—there is no upper limit. Even though there is an infinite number of possible values, this causes no real problems because the probabili- ties of all sufficiently large values are essentially 0.

The Poisson distribution is characterized by a single parameter, usually labeled l (Greek lambda), which must be positive. By adjusting the value of l, we are able to produce different Poisson distributions, all of which have the same basic shape as in Figure 5.21. That is, they first increase and then decrease. It turns out that l is easy to interpret. It is both the mean and the variance of the Poisson distribution. Therefore, the standard deviation is !l.

Figure 5.21 Typical Poisson Distribution

Typical Examples of the Poisson Distribution

• A bank manager is studying the arrival pattern to the bank. The events are customer arrivals, the number of arrivals in an hour is Poisson distributed, and l represents the expected number of arrivals per hour.

• An engineer is interested in the lifetime of a type of battery. A device that uses this type of battery is operated continuously. When the first battery fails, it is replaced by a sec- ond; when the second fails, it is replaced by a third, and so on. The events are battery failures, the number of failures that occur in a month is Poisson distributed, and l rep- resents the expected number of failures per month.

• A retailer is interested in the number of customers who order a particular product in a week. Then the events are customer orders for the product, the number of customer orders in a week is Poisson distributed, and l is the expected number of orders per week.

• In a quality control setting, the Poisson distribution is often relevant for describing the number of defects in some unit of space. For example, when paint is applied to the body of a new car, any minor blemish is considered a defect. Then the number of defects on the hood, say, might be Poisson distributed. In this case, l is the expected number of defects per hood.

These examples are representative of the many situations where the Poisson distribution has been applied. The parameter l is often called a rate—arrivals per hour, failures per month, and so on. If the unit of time (or space) is changed, the rate must be modified

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accordingly. For example, if the number of arrivals to a bank in a single hour is Poisson distributed with rate l 5 30, then the number of arrivals in a half-hour period is Poisson distributed with rate l 5 15.

You can use Excel to calculate Poisson probabilities much as you did with binomial probabilities. The relevant function is the POISSON function. It takes the form

=POISSON.DIST (k,l,cum)

The third argument cum works exactly as in the binomial case. If it is FALSE (or 0), the function returns P1X 5 k2; if it is TRUE (or 1), the function returns P1X # k2 . As examples, if l 5 5 5POISSON.DIST17,5,FALSE2 returns the probability of exactly 7, 5POISSON.DIST17,5,TRUE2 returns the probability of less than or equal to 7, and 5 1-POISSON.DIST13,5,TRUE2 returns the probability of greater than 3.

The next example shows how a manager or consultant could use the Poisson distribution.

EXAMPLE

5.13 MANAGING TV INVENTORY AT KRIEGLAND Kriegland is a department store that sells various brands of plasma screen TVs. One of the manager’s biggest problems is to decide on an appropriate inventory policy for stocking TVs. He wants to have enough in stock so that customers receive their requests right away, but he does not want to tie up too much money in inventory that sits on the storeroom floor.

Most of the difficulty results from the unpredictability of customer demand. If this demand were constant and known, the manager could decide on an appropriate inventory policy fairly easily. But the demand varies widely from month to month in a random manner. All the manager knows is that the historical average demand per month is approximately 17. Therefore, he decides to call in a consultant. The consultant immediately suggests using a probability model. Specifically, she attempts to find the probability distribution of demand in a typical month. How might she proceed?

Objective To model the probability distribution of monthly demand for plasma screen TVs with a particular Poisson distribution.

Solution Let X be the demand in a typical month. The consultant knows that there are many possible values of X. For example, if his- torical records show that monthly demands have always been between 0 and 40, the consultant knows that almost all of the probability should be assigned to the values 0 through 40. However, she does not relish the thought of finding 41 probabilities, P1X 5 02 through P1X 5 402, that sum to 1 and reflect historical frequencies. Instead, she discovers from the manager that the histogram of demands from previous months is shaped much like the graph in Figure 5.21. That is, it rises to some peak and then falls.

Knowing that a Poisson distribution has this same basic shape, the consultant decides to model the monthly demand with a Poisson distribution. To choose a particular Poisson distribution, all she has to do is choose a value of l, the mean demand per month. Because the historical average is approximately 17, she chooses l 5 17. Now she can test the Poisson model by calculating probabilities of various events and asking the manager whether these probabilities are reasonable approximations to reality.

For example, the Poisson probability that monthly demand is less than or equal to 20, P1X # 202 , is 0.805 [using the Excel function POISSON.DIST 120,17,TRUE2], and the probability that demand is between 10 and 15 inclusive, P110 # X # 152 , is 0.345 [using POISSON.DIST115,17,TRUE2-POISSON.DIST19,17,TRUE2]. Figure 5.22 illustrates various probability calculations and shows the graph of the individual Poisson probabilities. (See the file Poisson Demand Distribution Finished.xlsx.)

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5-6 The Poisson and Exponential Distributions 2 2 9

If the manager believes that these probabilities and other similar probabilities are reasonable, then the statistical part of the consultant’s job is finished. Otherwise, she must try a different Poisson distribution—a different value of l—or perhaps a different type of distribution altogether.

Value

8 0 00719

8 0 007

17 0 096

1 JIHGFEDCBA

Poisson distribution for monthly demand 2 3 4 5 6 7 8 9

Range name used: Mean monthly demand Mean =$B$3

Representative probability calculations Less than or equal to 20 0.805 =POISSON.DIST(20,Mean,TRUE) Between 10 and 15 (inclusive) 0.345 =POISSON.DIST(15,Mean,TRUE)-POISSON.DIST(9,Mean,TRUE)

Individual probabilities Value Probability

11 12 13 14 15 16 17 18

0 0.000 =POISSON.DIST(A11,Mean,FALSE)

7 0.003

20 21 22 23 24 25 26 27

10 0.023

28 29 30 31

1 0.000 copy down 2 0.000 3 0.000 4 0.000 5 0.000 6 0.001

9 0.014

11 0.036 12 0.050 13 0.066 14 0.080 15 0.091 16 0.096 17 0.096 18 0.091 19 0.081 20 0.069

Poisson Distribution with Mean 17 0.120

0.100

0.080

0.060

0.040

0.020

0.000 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40

Figure 5.22 Poisson Calculations for TV Example

5-6b The Exponential Distribution Suppose that a bank manager is studying the pattern of customer arrivals at her branch location. As indicated previously in this section, the number of arrivals in an hour at a facility such as a bank is often well described by a Poisson distribution with parameter l, where l represents the expected number of arrivals per hour. An alternative way to view the uncertainty in the arrival process is to consider the times between customer arrivals. The most common probability distribution used to model these times, often called interar- rival times, is the exponential distribution.

In general, the continuous random variable X has an exponential distribution with parameter l (with l 7 0) if the density function of X has the form f1x2 5 le2 lx for x 7 0. This density function has the shape shown in Figure 5.23. Because this density function decreases continuously from left to right, its most likely value is x 5 0. Alternatively, if you collect many observations from an exponential distribution and draw a histogram of the observed values, you should expect it to resemble the smooth curve shown in Figure 5.23, with the tallest bars to the left. The mean and standard deviation of this dis- tribution are easy to remember. They are both equal to the reciprocal of the parameter l. For example, an exponential distribution with parameter l 5 0.1 has mean and standard deviation both equal to 10.

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As with the normal distribution, you usually want probabilities to the left or right of a given value. For any exponential distribution, the probability to the left of a given value x 7 0 can be calculated with Excel’s EXPON.DIST function. This function takes the form

=EXPON.DIST(x,l,TRUE)

For example, if x 5 0.5 and l 5 5 (so that the mean equals 1>5 5 0.2), the probabil- ity of being less than 0.5 can be found with the formula

=EXPON.DIST(0.5,5,TRUE)

This returns the probability 0.918. Of course, the probability of being greater than 0.5 is then 1 2 0.918 5 0.082.

Returning to the bank manager’s analysis of customer arrival data, when the times between arrivals are exponentially distributed, you sometimes hear that “arrivals occur according to a Poisson process.” This is because there is a close relationship between the exponential distribution, which measures times between events such as arrivals, and the Poisson distribution, which counts the number of events in a certain length of time. The details of this relationship are beyond the level of this book, so we will not explore the topic further. But if you hear, for example, that customers arrive at a facility accord- ing to a Poisson process at the rate of six per hour, then the corresponding times between arrivals are exponentially distributed with mean 1/6 hour.

Figure 5.23 Exponential Density Function

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 44. The annual number of industrial accidents occurring in

a particular manufacturing plant is known to follow a Poisson distribution with mean 12. a. What is the probability of observing exactly 12 acci-

dents during the coming year? b. What is the probability of observing no more than 12

accidents during the coming year? c. What is the probability of observing at least 15 acci-

dents during the coming year? d. What is the probability of observing between 10 and

15 accidents (inclusive) during the coming year?

e. Find the smallest integer k such that we can be at least 99% sure that the annual number of accidents occur- ring will be less than k.

45. Suppose the number of baskets scored by the Indiana University basketball team in one minute follows a Poisson distribution with l 5 1.5. In a 10-minute span of time, what is the probability that Indiana University scores exactly 20 baskets; at most 20 baskets? (Use the fact that if the rate per minute is l, then the rate in t minutes is lt.)

46. Suppose that the times between arrivals at a bank during the peak period of the day are exponentially distrib- uted with a mean of 45 seconds. If you just observed an arrival, what is the probability that you will need to wait for more than a minute before observing the next arrival? What is the probability you will need to wait at least two minutes?

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5-7 Conclusion 2 3 1

Level B 47. Consider a Poisson random variable X with parameter

l 5 2. a. Find the probability that X is within one standard

deviation of its mean. b. Find the probability that X is within two standard

deviations of its mean. c. Find the probability that X is within three standard

deviations of its mean. d. Do the empirical rules we learned previously seem to

be applicable in working with the Poisson distribution where l 5 2? Explain why or why not.

e. Repeat parts a through d for the case of a Poisson ran- dom variable where l 5 20.

48. Based on historical data, the probability that a major league pitcher pitches a no-hitter in a game is about 1/1300. a. Use the binomial distribution to determine the proba-

bility that in 650 games 0, 1, 2, or 3 no-hitters will be pitched. (Find the separate probabilities of these four events.)

b. Repeat part a using the Poisson approximation to the binomial. This approximation says that if n is large and p is small, a binomial distribution with parame- ters n and p is approximately the same as a Poisson distribution with l 5 np.

5-7 Conclusion We have covered a lot of ground in this chapter, and much of the material, especially that on the normal distribution, will be used in later chapters. The normal distribution is the cornerstone for much of statistical theory. As you will see in later chapters on statistical inference and regression, an assumption of normality is behind most of the procedures we use. Therefore, it is important for you to understand the properties of the normal distribution and how to work with it in Excel. The binomial, Pois- son, and exponential distributions, although not used as frequently as the normal distribution in this book, are also extremely important. The examples we have discussed indicate how these distributions can be used in a variety of business situations. Finally, whenever a situation involves uncertainty, the rules of probability are relevant. Although we haven’t discussed the math- ematical details of probability in much depth, you should at least be familiar with the basic rules of probability discussed here.

Summary of Key Terms TERM EXPLANATION EXCEL PAGE EQUATION

Random variable Associates a numeric value with each possible outcome in a situation involving uncertainty

140

Probability A number between 0 and 1 that measures the likeli- hood that some event will occur

142

Rule of complements The probability of any event and the probability of its complement sum to 1

Basic formulas

142 5.1

Mutually exclusive events Events where only one of them can occur 142

Exhaustive events Events where at least one of them must occur 143

Addition rule for mutually exclusive events

The probability that at least one of a set of mutually exclusive events will occur is the sum of their probabilities

Basic formulas

143 5.2

Conditional probability formula

Updates the probability of an event, given the knowl- edge that another event has occurred

Basic formulas

144 5.3

Multiplication rule Formula for the probability that two events both occur Basic formulas

144 5.4

Probability tree A graphical representation of how events occur through time, useful for calculating probabilities of multiple events

145

Probabilistically indepen- dent events

Events where knowledge that one of them has occurred is of no value in assessing the probability that the other will occur

146 5.5

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TERM EXPLANATION EXCEL PAGE EQUATION

Relative frequency The proportion of times the event occurs out of the number of times a random experiment is performed

147

Cumulative probability “Less than or equal to” probabilities associated with a random variable

150

Mean (or expected value) of a probability distribution

A measure of central tendency—the weighted sum of the possible values, weighted by their probabilities

Basic formulas

151 5.6

Variance of a probability distribution

A measure of variability: the weighted sum of the squared deviations of the possible values from the mean, weighted by the probabilities

Basic formulas

151 5.7, 5.9

Standard deviation of a probability distribution

A measure of variability: the square root of the variance Basic formulas

151 5.8

Density function Specifies the probability distribution of a continuous random variable

168

Normal distribution A continuous distribution with possible values ranging over the entire number line; its density function is a symmetric bell-shaped curve

169 5.1

Standardizing a normal random variable

Transforms any normal distribution with mean m and standard deviation s to the standard normal distribution with mean 0 and standard deviation 1

170 5.2

Normal calculations in Excel

Useful for finding probabilities and percentiles for non- standard and standard normal distributions

NORM.DIST, NORM.S.DIST, NORM.INV, NORM.S.INV

175

Empirical rules for normal distribution

About 68% of the data fall within one standard deviation of the mean, about 95% of the data fall within two stan- dard deviations of the mean, and almost all fall within three standard deviations of the mean

177

Binomial distribution The distribution of the number of successes in n inde- pendent, identical trials, where each trial has probability p of success

BINOM.DIST BINOM.INV

190

Mean and standard deviation of a binomial distribution

The mean and standard deviation of a binomial distri- bution with parameters n and p are np and !np 11 2 p2, respectively

193 5.3, 5.4

Sampling without replacement

Sampling where no member of the population can be sampled more than once

194

Hypergeometric distribution

Relevant when sampling without replacement, espe- cially when the fraction of the population sampled is large

HYPGEOM. DIST

Sampling with replacement

Sampling where any member of the population can be sampled more than once

194

Normal approximation to the binomial distribution

If np 7 5 and n11 2 p2 7 5, the binomial distribution can be approximated well by a normal distribution with mean np and standard deviation !np11 2 p2

194

Poisson distribution A discrete probability distribution that often describes the number of events occurring within a specified period of time or space; mean and variance both equal the parameter l

POISSON.DIST 207

Exponential distribution A continuous probability distribution useful for measur- ing times between events, such as customer arrivals to a service facility; mean and standard deviation both equal the reciprocal of the parameter

EXPON.DIST 210

Key Terms (continued)

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5-7 Conclusion 2 3 3

Problems

Conceptual Questions C.1. Suppose that you want to find the probability that

event A or event B will occur. If these two events are not mutually exclusive, explain how you would proceed.

C.2. “If two events are mutually exclusive, they must not be independent events.” Is this statement true or false? Explain your choice.

C.3. Is the number of passengers who show up for a particu- lar commercial airline flight a discrete or a continuous random variable? Is the time between flight arrivals at a major airport a discrete or a continuous random vari- able? Explain your answers.

C.4. Suppose that officials in the federal government are trying to determine the likelihood of a major small- pox epidemic in the United States within the next 12 months. Is this an example of an objective probability or a subjective probability? How might the officials assess this probability?

C.5. Consider the statement, “When there are a finite num- ber of outcomes, then all probability is just a matter of counting. Specifically, if n of the outcomes are favor- able to some event E, and there are N outcomes total, then the probability of E is n/N .” Is this statement always true? Is it always false?

C.6. If there is uncertainty about some monetary outcome and you are concerned about return and risk, then all you need to see are the mean and standard deviation. The entire distribution provides no extra useful infor- mation. Do you agree or disagree? Provide an example to back up your argument.

C.7. Choose at least one uncertain quantity of interest to you. For example, you might choose the highest price of gas between now and the end of the year, the high- est point the Dow Jones Industrial Average will reach between now and the end of the year, the number of majors Tiger Woods will win in his career, and so on. Using all of the information and insight you have, assess the probability distribution of this uncertain quantity. Is there one “right answer?”

C.8. Historically, the most popular measure of variability has been the standard deviation, the square root of the weighted sum of squared deviations from the mean, weighted by their probabilities. Suppose analysts had always used an alternative measure of variability, the weighted sum of the absolute deviations from the mean, again weighted by their probabilities. Do you think this would have made a big difference in the the- ory and practice of probability and statistics?

C.9. Suppose a person flips a coin, but before you can see the result, the person puts her hand over the coin. At this point, does it make sense to talk about the probability

that the result is heads? Is this any different from the probability of heads before the coin was flipped?

C.10. Consider an event that will either occur or not. For example, the event might be that California will expe- rience a major earthquake in the next five years. You let p be the probability that the event will occur. Does it make any sense to have a probability distribution of p? Why or why not? If so, what might this distribution look like? How would you interpret it?

C.11. Suppose a couple is planning to have two children. Let B1 be the event that the first child is a boy, and let B2 be the event that the second child is a boy. You and your friend get into an argument about whether B1 and B2 are independent events. You think they are indepen- dent and your friend thinks they aren’t. Which of you is correct? How could you settle the argument?

C.12 For each of the following uncertain quantities, discuss whether it is reasonable to assume that the probabil- ity distribution of the quantity is normal. If the answer isn’t obvious, discuss how you could discover whether a normal distribution is reasonable. a. The change in the Dow Jones Industrial Average

between now and a year from now b. The length of time (in hours) a battery that is in con-

tinuous use lasts c. The time between two successive arrivals to a bank d. The time it takes a bank teller to service a random

customer e. The length (in yards) of a typical drive on a par 5 by

Phil Michelson f. The amount of snowfall (in inches) in a typical win-

ter in Minneapolis g. The average height (in inches) of all boys in a ran-

domly selected seventh-grade middle school class h. Your bonus from finishing a project, where your

bonus is $1000 per day under the deadline if the proj- ect is completed before the deadline, your bonus is $500 if the project is completed right on the dead- line, and your bonus is $0 if the project is completed after the deadline

i. Your gain on a call option on a stock, where you gain nothing if the price of the stock a month from now is less than or equal to $50 and you gain 1P 2 502 dollars if the price P a month from now is greater than $50

C.13 For each of the following uncertain quantities, discuss whether it is reasonable to assume that the probabil- ity distribution of the quantity is binomial. If you think it is, what are the parameters n and p? If you think it isn’t, explain your reasoning. a. The number of wins the Boston Red Sox baseball

team has next year in its 81 home games b. The number of free throws Kobe Bryant misses in

his next 250 attempts c. The number of free throws it takes Kobe Bryant to

achieve 100 successes

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d. The number out of 1000 randomly selected custom- ers in a supermarket who have a bill of at least $150

e. The number of trading days in a typical year where Microsoft’s stock price increases

f. The number of spades you get in a 13-card hand from a well-shuffled 52-card deck

g. The number of adjacent 15-minute segments during a typical Friday where at least 10 customers enter a McDonald’s restaurant

h. The number of pages in a 500-page book with at least one misprint on the page

C.14 One disadvantage of a normal distribution is that there is always some probability that a quantity is negative, even when this makes no sense for the uncertain quan- tity. For example, the time a light bulb lasts cannot be negative. In any particular situation, how would you decide whether you could ignore this disadvantage for all practical purposes?

C.15 For real applications, the normal distribution has two potential drawbacks: (1) it can be negative, and (2) it isn’t symmetric. Choose some continuous random numeric outcomes of interest to you. Are either poten- tial drawbacks really drawbacks for your random out- comes? If so, which is the more serious drawback?

C.16 Many basketball players and fans believe strongly in the “hot hand.” That is, they believe that players tend to shoot in streaks, either makes or misses. If this is the case, why does the binomial distribution not apply, at least not exactly, to the number of makes in a given number of shots? Which assumption of the binomial model is violated, the independence of successive shots or the con- stant probability of success on each shot? Or can you tell?

Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 49. A business manager who needs to make many phone

calls has estimated that when she calls a client, the prob- ability that she will reach the client right away is 60%. If she does not reach the client on the first call, the proba- bility that she will reach the client with a subsequent call in the next hour is 20%. a. Find the probability that the manager reaches her

client in two or fewer calls. b. Find the probability that the manager reaches her

client on the second call but not on the first call. c. Find the probability that the manager is unsuccessful

on two consecutive calls. 50. Suppose that a marketing research firm sends

questionnaires to two different companies. Based on historical evidence, the marketing research firm believes that each company, independently of the other, will return the questionnaire with probability 0.40. a. What is the probability that both questionnaires are

returned?

b. What is the probability that neither of the question- naires is returned?

c. Now, suppose that this marketing research firm sends questionnaires to ten different companies. Assum- ing that each company, independently of the others, returns its completed questionnaire with probability 0.40, how do your answers to parts a and b change?

51. Based on past sales experience, an appliance store stocks five window air conditioner units for the coming week. No orders for additional air conditioners will be made until next week. The weekly consumer demand for this type of appliance has the probability distribution given in the file P05_51.xlsx. a. Let X be the number of window air conditioner units

left at the end of the week (if any), and let Y be the number of special stockout orders required (if any), assuming that a special stockout order is required each time there is a demand and no unit is available in stock. Find the probability distributions of X and Y.

b. Find the expected value of X and the expected value of Y.

c. Assume that this appliance store makes a $60 profit on each air conditioner sold from the weekly available stock, but the store loses $20 for each unit sold on a spe- cial stockout order basis. Let Z be the profit that the store earns in the coming week from the sale of window air conditioners. Find the probability distribution of Z.

d. Find the expected value of Z. 52. A roulette wheel contains the numbers 0, 00, and 1 to

36. If you bet $1 on a single number coming up, you earn $35 if the number comes up and lose $1 otherwise. Find the mean and standard deviation of your winnings on a single bet. Then find the mean and standard devia- tion of your net winnings if you make 100 bets. You can assume (realistically) that the results of the 100 spins are independent. Finally, provide an interval such that you are 95% sure your net winnings from 100 bets will be inside this interval.

53. You are involved in a risky business venture where three outcomes are possible: (1) you will lose not only your initial investment ($5000) but an additional $3000; (2) you will just make back your initial investment (for a net gain of $0); or (3) you will make back your initial investment plus an extra $10,000.

The probability of (1) is half as large as the probability of (2), and the probability of (3) is one-third as large as the probability of (2). a. Find the individual probabilities of (1), (2), and (3).

(They should sum to 1.) b. Find the expected value and standard deviation of

your net gain (or loss) from this venture. 54. Suppose the annual return on XYZ stock follows a nor-

mal distribution with mean 12% and standard deviation 30%. a. What is the probability that XYZ’s value will decrease

during a year?

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5-7 Conclusion 2 3 5

b. What is the probability that the return on XYZ during a year will be at least 20%?

c. What is the probability that the return on XYZ during a year will be between 26% and 9%?

d. There is a 5% chance that the return on XYZ during a year will be greater than what value?

e. There is a 1% chance that the return on XYZ during a year will be less than what value?

f. There is a 95% chance that the return on XYZ during a year will be between which two values (equidistant from the mean)?

55. A family is considering a move from a midwestern city to a city in California. The distribution of housing costs where the family currently lives is normal, with mean $105,000 and standard deviation $18,200. The distri- bution of housing costs in the California city is normal with mean $235,000 and standard deviation $30,400. The family’s current house is valued at $110,000. a. What percentage of houses in the family’s current city

cost less than theirs? b. If the family buys a $200,000 house in the new city,

what percentage of houses there will cost less than theirs?

c. What price house will the family need to buy to be in the same percentile (of housing costs) in the new city as they are in the current city?

56. The number of traffic fatalities in a typical month in a given state has a normal distribution with mean 125 and standard deviation 31. a. If a person in the highway department claims that

there will be at least m fatalities in the next month with probability 0.95, what value of m makes this claim true?

b. If the claim is that there will be no more than n fatal- ities in the next month with probability 0.98, what value of n makes this claim true?

57. It can be shown that a sum of normally distributed ran- dom variables is also normally distributed. Do all func- tions of normal random variables lead to normal random variables? Consider the following. SuperDrugs is a chain of drugstores with three similar-size stores in a given city. The sales in a given week for any of these stores is normally distributed with mean $15,000 and standard deviation $3000. At the end of each week, the sales fig- ure for the store with the largest sales among the three stores is recorded. Is this maximum value normally dis- tributed? To answer this question, simulate a weekly sales figure at each of the three stores and calculate the maximum. Then replicate this maximum 500 times and create a histogram of the 500 maximum values. Does it appear to be normally shaped? Whatever this distribu- tion looks like, use your simulated values to estimate the mean and standard deviation of the maximum.

58. In the financial world, there are many types of com- plex instruments called derivatives that derive their value from the value of an underlying asset. Consider

the following simple derivative. A stock’s current price is $80 per share. You purchase a derivative whose value to you becomes known a month from now. Specifi- cally, let P be the price of the stock in a month. If P is between $75 and $85, the derivative is worth noth- ing to you. If P is less than $75, the derivative results in a loss of 100*175 2 P2 dollars to you. (The factor of 100 is because many derivatives involve 100 shares.) If P is greater than $85, the derivative results in a gain of 100*1P 2 852 dollars to you. Assume that the distri- bution of the change in the stock price from now to a month from now is normally distributed with mean $1 and standard deviation $8. Let P (big loss) be the proba- bility that you lose at least $1000 (that is, the price falls below $65), and let P (big gain) be the probability that you gain at least $1000 (that is, the price rises above $95). Find these two probabilities. How do they com- pare to one another?

Level B 59. Equation (5.7) for variance indicates exactly what vari-

ance is: the weighted average of squared deviations from the mean, weighted by the probabilities. However, the computing formula for variance, Equation (5.9), is more convenient for spreadsheet calculations. Show algebra- ically that the two formulas are equivalent.

60. The basic game of craps works as follows. You throw two dice. If the sum of the two faces showing up is 7 or 11, you win and the game is over. If the sum is 2, 3, or 12, you lose and the game is over. If the sum is anything else (4, 5, 6, 8, 9, or 10), that value becomes your “point.” You then keep throwing the dice until the sum matches your point or equals 7. If your point occurs first, you win and the game is over. If 7 occurs first, you lose and the game is over. What is the probability that you win the game?

61. Consider an individual selected at random from a sample of 750 married women (see the data in the file P05_61. xlsx) in answering each of the following questions. a. What is the probability that this woman does not work

outside the home, given that she has at least one child? b. What is the probability that this woman has no chil-

dren, given that she works part time? c. What is the probability that this woman has at least

two children, given that she does not work full time? 62. Suppose that 8% of all managers in a given company

are African American, 13% are women, and 17% have earned an MBA degree from a top-10 graduate business school. Let A, B, and C be, respectively, the events that a randomly selected individual from this population is African American, is a woman, and has earned an MBA from a top-10 graduate business school. a. Do you believe that A, B, and C are independent

events? Explain why or why not.

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b. Assuming that A, B, and C are independent events, find the probability that a randomly selected manager from this company is a white male and has earned an MBA degree from a top-10 graduate business school.

c. If A, B, and C are not independent events, can you calculate the probability requested in part b from the information given? What further information would you need?

63. Two gamblers play a version of roulette with a wheel as shown in the file P05_63.xlsx. Each gambler places four bets, but their strategies are different, as explained below. For each gambler, use the rules of probability to find the distribution of their net winnings after four bets. Then find the mean and standard deviation of their net winnings. The file gets you started. a. Player 1 always bets on red. On each bet, he either

wins or loses what he bets. His first bet is for $10. From then on, he bets $10 following a win, and he doubles his bet after a loss. (This is called a martingale strategy and is used frequently at casinos.) For example, if he spins red, red, not red, and not red, his bets are for $10, $10, $10, and $20, and he has a net loss of $10. Or if he spins not red, not red, not red, and red, then his bets are for $10, $20, $40, and $80, and he has a net gain of $10.

b. Player 2 always bets on black and green. On each bet, he places $10 on black and $2 on green. If red occurs, he loses all $12. If black occurs, he wins a net $8 ($10 gain on black, $2 loss on green). If green occurs, he wins a net $50 ($10 loss on black, $60 gain on green).

64. Suppose the New York Yankees and Philadelphia Phil- lies (two Major League Baseball teams) are playing a best-of-three series. The first team to win two games is the winner of the series, and the series ends as soon as one team has won two games. The first game is played in New York, the second game is in Phil- adelphia, and if necessary the third game is in New York. The probability that the Yankees win a game in their home park is 0.55. The probability that the Phil- lies win a game in their home park is 0.53. You can assume that the outcomes of the games are probabilis- tically independent. a. Find the probability that the Yankees win the series. b. Suppose you are a Yankees fan, so you place a bet

on each game played where you win $100 if the Yankees win the game and you lose $105 if the Yankees lose the game. Find the distribution of your net winnings. Then find the mean and standard deviation of this distribution. Is this betting strategy favorable to you?

c. Repeat part a, but assume that the games are played in Philadelphia, then New York, then Philadelphia. How much does this “home field advantage” help the Phillies?

d. Repeat part a, but now assume that the series is a best- of-five series, where the first team that wins three games wins the series. Assume that games alternate between New York and Philadelphia, with the first game in New York.

65. Have you ever watched the odds at a horse race? You might hear that the odds against a given horse win- ning are 9 to 1, meaning that the horse has probability 1> 11 1 92 5 1>10 of winning. However, these odds, after being converted to probabilities, typically sum to something greater than one. Why is this? Suppose you place a bet of $10 on this horse. It seems that it is a fair bet if you lose your $10 if the horse loses, but you win $90 if the horse wins. However, argue why this isn’t really fair to you, that is, argue why your expected win- nings are negative.

66. When you sum 30 or more independent random vari- ables, the sum of the random variables will usually be approximately normally distributed, even if each individual random variable is not normally distributed. Use this fact to estimate the probability that a casino will be behind after 90,000 roulette bets, given that it wins $1 or loses $35 on each bet with probabilities 37/38 and 1/38.

67. Many companies use sampling to determine whether a batch should be accepted. An 1n, c2 sampling plan con- sists of inspecting n randomly chosen items from a batch and accepting the batch if c or fewer sampled items are defective. Suppose a company uses a 1100, 52 sampling plan to determine whether a batch of 10,000 computer chips is acceptable. a. The “producer’s risk” of a sampling plan is the prob-

ability that an acceptable batch will be rejected by the sampling plan. Suppose the customer considers a batch with 3% defectives acceptable. What is the pro- ducer’s risk for this sampling plan?

b. The “consumer’s risk” of a sampling plan is the prob- ability that an unacceptable batch will be accepted by the sampling plan. Our customer says that a batch with 9% defectives is unacceptable. What is the con- sumer’s risk for this sampling plan?

68. Suppose that if a presidential election were held today, 53% of all voters would vote for candidate Smith over candidate Jones. (You can substitute the names of the most recent presidential candidates.) This problem shows that even if there are 100 million voters, a sample of several thousand is enough to determine the outcome, even in a fairly close election. a. If 1500 voters are sampled randomly, what is the

probability that the sample will indicate (correctly) that Smith is preferred to Jones?

b. If 6000 voters are sampled randomly, what is the probability that the sample will indicate (correctly) that Smith is preferred to Jones?

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69. A company assembles a large part by joining two smaller parts together. Assume that the smaller parts are normally distributed with a mean length of 1 inch and a standard deviation of 0.01 inch. a. What fraction of the larger parts are longer than 2.05

inches? b. What fraction of the larger parts are between 1.96

inches and 2.02 inches long? 70. (Suggested by Sam Kaufmann, Indiana University MBA,

who has run Harrah’s Lake Tahoe Casino.) A high roller has come to the casino to play 300 games of craps. For each game of craps played there is a 0.493 probability that the high roller will win $1 and a 0.507 probabil- ity that the high roller will lose $1. After 300 games of craps, what is the probability that the casino will be behind more than $10?

71. (Suggested by Sam Kaufmann, Indiana University MBA, who has run Harrah’s Lake Tahoe Casino.) A high roller comes to the casino intending to play 500 hands of blackjack for $1 a hand. On each hand, the high roller will win $1 with probability 0.48 and lose $1 with prob- ability 0.52. After the 500 hands, what is the probability that the casino has lost more than $40?

72. The weekly demand for TVs at Lowland Appliance is normally distributed with mean 400 and standard devia- tion 100. Each time an order for TVs is placed, it arrives exactly four weeks later. That is, TV orders have a four- week lead time. Lowland doesn’t want to run out of TVs during any more than 1% of all lead times. How low should Lowland let its TV inventory drop before it places an order for more TVs? (Hint: How many stan- dard deviations above the mean lead-time demand must the reorder point be for there to be a 1% chance of a stockout during the lead time?)

73. An elevator rail is assumed to meet specifications if its diameter is between 0.98 and 1.01 inches. Each year a company produces 100, 000 elevator rails. For a cost of $10/s2 per year the company can rent a machine that produces elevator rails whose diame- ters have a standard deviation of s. (The idea is that the company must pay more for a smaller variance.) Each such machine will produce rails having a mean diameter of one inch. Any rail that does not meet spec- ifications must be reworked at a cost of $12. Assume that the diameter of an elevator rail follows a normal distribution. a. What standard deviation (within 0.001 inch) mini-

mizes the annual cost of producing elevator rails? You do not need to try standard deviations in excess of 0.02 inch.

b. For your answer in part a, one elevator rail in 1000 will be at least how many inches in diameter?

74. What caused the crash of TWA Flight 800 in 1996? Physics professors Hailey and Helfand of Columbia University

believe there is a reasonable possibility that a meteor hit Flight 800. They reason as follows. On a given day, 3000 meteors of a size large enough to destroy an airplane hit the earth’s atmosphere. Approximately 50,000 flights per day, averaging two hours in length, have been flown from 1950 to 1996. This means that at any given point in time, planes in flight cover approximately two-billionths of the world’s atmosphere. Determine the probability that at least one plane in the last 47 years has been downed by a meteor. (Hint: Use the Poisson approximation to the binomial. This approximation says that if n is large and p is small, a binomial distribution with parameters n and p is approximately Poisson distributed with l 5 np.)

75. In the decade 1982 through 1991, 10 employees work- ing at the Amoco Company chemical research center were stricken with brain tumors. The average employ- ment at the center was 2000 employees. Nationwide, the average incidence of brain tumors in a single year is 20 per 100,000 people. If the incidence of brain tumors at the Amoco chemical research center were the same as the nationwide incidence, what is the probability that at least 10 brain tumors would have been observed among Amoco workers during the decade 1982 through 1991? What do you conclude from your analysis? (Source: AP wire service report, March 12, 1994)

76. Claims arrive at random times to an insurance company. The daily amount of claims is normally distributed with mean $1570 and standard deviation $450. Total claims on different days each have this distribution, and they are probabilistically independent of one another. a. Find the probability that the amount of total claims

over a period of 100 days is at least $150,000. b. If the company receives premiums totaling $165,000,

find the probability that the company will net at least $10,000 for the 100-day period.

77. Your company is running an audit on Sleaze Company. Because Sleaze has a bad habit of overcharging its cus- tomers, the focus of your audit is on checking whether the billing amounts on its invoices are correct. Assume that each invoice is for too high an amount with prob- ability 0.06 and for too low an amount with probabil- ity 0.01 (so that the probability of a correct billing is 0.93). Also, assume that the outcome for any invoice is probabilistically independent of the outcomes for other invoices. a. If you randomly sample 200 of Sleaze’s invoices,

what is the probability that you will find at least 15 invoices that overcharge the customer? What is the probability you won’t find any that undercharge the customer?

b. Find an integer k such that the probability is at least 0.99 that you will find at least k invoices that over- charge the customer. (Hint: Use trial and error with the BINOM.DIST function to find k.)

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78. Continuing the previous problem, suppose that when Sleaze overcharges a customer, the distribution of the amount overcharged (expressed as a percentage of the correct billing amount) is normally distributed with mean 15% and standard deviation 4%. a. What percentage of overbilled customers are charged

at least 10% more than they should pay? b. What percentage of all customers are charged at least

10% more than they should pay? c. If your auditing company samples 200 randomly cho-

sen invoices, what is the probability that it will find at least five where the customer was overcharged by at least 10%?

79. As any credit-granting agency knows, there are always some customers who default on credit charges. Typically, customers are grouped into relatively homogeneous categories, so that customers within any category have approximately the same chance of defaulting on their credit charges. Here we will look at one particular group of customers. We assume each of these customers has (1) probability 0.07 of defaulting on his or her current credit charges, and (2) total credit charges that are nor- mally distributed with mean $350 and standard devia- tion $100. We also assume that if a customer defaults, 20% of his or her charges can be recovered. The other 80% are written off as bad debt. a. What is the probability that a typical customer in this

group will default and produce a write-off of more than $250 in bad debt?

b. If there are 500 customers in this group, what are the mean and standard deviation of the number of cus- tomers who will meet the description in part a?

c. Again assuming there are 500 customers in this group, what is the probability that at least 25 of them will meet the description in part a?

d. Suppose now that nothing is recovered from a default—the whole amount is written off as bad debt. Show how to simulate the total amount of bad debt from 500 customers in just two cells, one with a binomial calculation, the other with a normal calculation.

80. The Excel functions discussed in this chapter are use- ful for solving a lot of probability problems, but there are other problems that, even though they are similar to normal or binomial problems, cannot be solved with these functions. In cases like this, simulation can often be used. Here are a couple of such problems for you to simulate. For each example, simulate 500 replications of the experiment. a. You observe a sequence of parts from a manufactur-

ing line. These parts use a component that is supplied

by one of two suppliers. Each part made with a com- ponent from supplier 1 works properly with prob- ability 0.95, and each part made with a component from supplier 2 works properly with probability 0.98. Assuming that 100 of these parts are made, 60 from supplier 1 and 40 from supplier 2, you want the probability that at least 97 of them work properly.

b. Here we look at a more generic example such as coin flipping. There is a sequence of trials where each trial is a success with probability p and a failure with prob- ability 1 2 p. A run is a sequence of consecutive suc- cesses or failures. For most of us, intuition says that there should not be long runs. Test this by finding the probability that there is at least one run of length at least six in a sequence of 15 trials. (The run could be of 0s or 1s.) You can use any value of p you like—or try different values of p.

81. In the game of soccer, players are sometimes awarded a penalty kick. The player who kicks places the ball 12 yards from the 24-foot-wide goal and attempts to kick it past the goalie into the net. (The goalie is the only defender.) The question is where the player should aim. Make the following assumptions. (1) The player’s kick is off target from where he aims, left or right, by a normally distributed amount with mean 0 and some standard devi- ation. (2) The goalie typically guesses left or right and dives in that direction at the moment the player kicks. If the goalie guesses wrong, he won’t block the kick, but if he guesses correctly, he will be able to block a kick that would have gone into the net as long as the kick is within a distance d from the middle of the goal. The goalie is equally likely to guess left or right. (3) The player never misses high, but he can miss to the right of the goal (if he aims to the right) or to the left (if he aims to the left). For reasonable values of the standard deviation and d, find the probability that the player makes a goal if he aims at a point t feet inside the goal. (By symmetry, you can assume he aims to the right, although the goalie doesn’t know this.) What value of t seems to maximize the prob- ability of making a goal?

82. In the 2012 Major League Baseball season, the Bal- timore Orioles were ahead after the 7th inning in 74 games, and they won all 74 games. Use an appro- priate model to explore how unusual such a streak is. Would you place it in the same category as the famous 56-game hitting streak (at least one hit per game) by Joe DiMaggio in 1941? Discuss the differences, including those that are caused by pressure.

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CASE 5.2 EuroWatch Company EuroWatch Company assembles expensive wristwatches and then sells them to retailers throughout Europe. The watches are assembled at a plant with two assembly lines. These lines are intended to be identical, but line 1 uses somewhat older equipment than line 2 and is typically less reliable. Histori- cal data have shown that each watch coming off line 1, inde- pendently of the others, is free of defects with probability 0.98. The similar probability for line 2 is 0.99. Each line produces 500 watches per hour. The production manager has asked you to answer the following questions.

1. She wants to know how many defect-free watches each line is likely to produce in a given hour. Specifically, find the smallest integer k (for each line separately) such that you can be 99% sure that the line will not produce more than k defective watches in a given hour.

2. EuroWatch currently has an order for 500 watches from an important customer. The company plans to fill this

order by packing slightly more than 500 watches, all from line 2, and sending this package off to the cus- tomer. Obviously, EuroWatch wants to send as few watches as possible, but it wants to be 99% sure that when the customer opens the package, there are at least 500 defect-free watches. How many watches should be packed?

3. EuroWatch has another order for 1000 watches. Now it plans to fill this order by packing slightly more than one hour’s production from each line. This package will contain the same number of watches from each line. As in the previous question, EuroWatch wants to send as few watches as possible, but it again wants to be 99% sure that when the customer opens the package, there are at least 1000 defect-free watches. The question of how many watches to pack is unfortunately quite diffi- cult because the total number of defect-free watches is not binomially distributed. (Why not?) Therefore, the

CASE 5.1 Simpson’s Paradox The results we obtain with conditional probabilities can be quite counterintuitive, even paradoxical. This case is similar to one described in an article by Blyth (1972), and is usually referred to as Simpson’s paradox. [Two other examples of Simpson’s paradox are described in articles by Westbrooke (1998) and Appleton et al. (1996).] Essentially, Simpson’s paradox says that even if one treatment has a better effect than another on each of two separate subpopulations, it can have a worse effect on the population as a whole.

Suppose that the population is the set of managers in a large company. We categorize the managers as those with an MBA degree (the Bs) and those without an MBA degree (the Bs). These categories are the two “treatment” groups. We also categorize the managers as those who were hired directly out of school by this company (the Cs) and those who worked with another company first (the Cs). These two categories form the two “subpopulations.” Finally, we use as a measure of effectiveness those managers who have been promoted within the past year (the As).

Assume the following conditional probabilities are given:

P1AuB and C2 5 0.10, P1AuB and C2 5 0.05 (5.16)

P1AuB and C2 5 0.35, P1AuB and C2 5 0.20 (5.17)

P1CuB2 5 0.90, P1CuB2 5 0.30 (5.18)

Each of these can be interpreted as a proportion. For exam- ple, the probability P1AuB and C2 implies that 10% of all

managers who have an MBA degree and were hired by the company directly out of school were promoted last year. Similar explanations hold for the other probabilities.

Joan Seymour, the head of personnel at this company, is trying to understand these figures. From the proba- bilities in Equation (5.16), she sees that among the sub- population of workers hired directly out of school, those with an MBA degree are twice as likely to be promoted as those without an MBA degree. Similarly, from the prob- abilities in Equation (5.17), she sees that among the sub- population of workers hired after working with another company, those with an MBA degree are almost twice as likely to be promoted as those without an MBA degree. The information provided by the probabilities in Equation (5.18) is somewhat different. From these, she sees that employees with MBA degrees are three times as likely as those without MBA degrees to have been hired directly out of school.

Joan can hardly believe it when a whiz-kid analyst uses these probabilities to show—correctly—that

P1AuB2 5 0.125, P1AuB2 5 0.155 (5.19)

In words, those employees without MBA degrees are more likely to be promoted than those with MBA degrees. This appears to go directly against the evidence in Equations (5.16) and (5.17), both of which imply that MBAs have an advantage in being promoted. Can you derive the proba- bilities in Equation (5.19)? Can you shed any light on this “paradox”?

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manager asks you to solve the problem with simulation (and some trial and error). (Hint: It turns out that it is much faster to simulate small numbers than large num- bers, so simulate the number of watches with defects, not the number without defects.)

4. Finally, EuroWatch has a third order for 100 watches. The customer has agreed to pay $50,000 for the order— that is, $500 per watch. If EuroWatch sends more than 100 watches to the customer, its revenue doesn’t increase; it can never exceed $50,000. Its unit cost of producing a watch is $450, regardless of which line it is assembled on. The order will be filled entirely from a single line, and EuroWatch plans to send slightly more than 100 watches to the customer.

If the customer opens the shipment and finds that there are fewer than 100 defect-free watches (which we assume the customer has the ability to do), then he will pay only for the defect-free watches—EuroWatch’s revenue will decrease by $500 per watch short of the 100 required—and on top of this, EuroWatch will be

required to make up the difference at an expedited cost of $1000 per watch. The customer won’t pay a dime for these expedited watches. (If expediting is required, EuroWatch will make sure that the expedited watches are defect-free. It doesn’t want to lose this customer entirely.)

You have been asked to develop a spreadsheet model to find EuroWatch’s expected profit for any number of watches it sends to the customer. You should develop it so that it responds correctly, regardless of which assem- bly line is used to fill the order and what the shipment quantity is. (Hints: Use the BINOM.DIST function, with last argument 0, to fill up a column of probabilities for each possible number of defective watches. Next to each of these, calculate EuroWatch’s profit. Then use a SUM- PRODUCT to obtain the expected profit. Finally, you can assume that EuroWatch will never send more than 110 watches. It turns out that this large a shipment is not even close to optimal.)

CASE 5.3 Cashing in on the Lottery Many states supplement their tax revenues with state-spon- sored lotteries. Most of them do so with a game called lotto. Although there are various versions of this game, they are all basically as follows. People purchase tickets that contain r distinct numbers from 1 to m, where r is generally 5 or 6 and m is generally around 50. For example, in Virginia, the state discussed in this case, r 5 6 and m 5 44. Each ticket costs $1, about 39 cents of which is allocated to the total jackpot.6 There is eventually a drawing of r 5 6 distinct numbers from the m 5 44 possible numbers. Any ticket that matches these 6 numbers wins the jackpot.

There are two interesting aspects of this game. First, the current jackpot includes not only the revenue from this round of ticket purchases but also any jackpots carried over from previous drawings because of no winning tickets. Therefore, the jackpot can build from one drawing to the next, and in celebrated cases it has become huge. Second, if there is more than one winning ticket—a distinct possibility—the winners share the jackpot equally. (This is called parimutuel betting.) So, for example, if the current jackpot is $9 million and there are three winning tickets, then each winner receives $3 million.

It can be shown that for Virginia’s choice of r and m, there are approximately 7 million possible tickets (7,059,052 to be exact). Therefore, any ticket has about one chance out

of 7 million of being a winner. That is, the probability of winning with a single ticket is p 5 1/7,059,052—not very good odds. If n people purchase tickets, then the number of winners is binomially distributed with parameters n and p. Because n is typically very large and p is small, the num- ber of winners has approximately a Poisson distribution with rate l 5 np. (This makes ensuing calculations somewhat easier.) For example, if 1 million tickets are purchased, then the number of winning tickets is approximately Poisson dis- tributed with l 5 1>7.

In 1992, an Australian syndicate purchased a huge num- ber of tickets in the Virginia lottery in an attempt to assure itself of purchasing a winner. It worked! Although the syn- dicate wasn’t able to purchase all 7 million possible tickets (it was about 1.5 million shy of this), it did purchase a win- ning ticket, and there were no other winners. Therefore, the syndicate won a 20-year income stream worth approximately $27 million, with a net present value of approximately $14 million. This made the syndicate a big profit over the cost of the tickets it purchased. Two questions come to mind: (1) Is this hogging of tickets unfair to the rest of the public? (2) Is it a wise strategy on the part of the syndicate (or did it just get lucky)?

To answer the first question, consider how the lottery changes for the general public with the addition of the syn- dicate. To be specific, suppose the syndicate can invest $7 million and obtain all of the possible tickets, making itself a sure winner. Also, suppose n people from the general public purchase tickets, each of which has 1 chance out of 7 mil- lion of being a winner. Finally, let R be the jackpot carried

6 Of the remaining 61 cents, the state takes about 50 cents. The other 11 cents is used to pay off lesser prize winners whose tickets match some, but not all, of the winning 6 numbers. To keep this case relatively simple, however, we ignore these lesser prizes and concentrate only on the jackpot.

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over from any previous lotteries. Then the total jackpot on this round will be 3R 1 0.39 17,000,000 1 n2 4 because 39 cents from every ticket goes toward the jackpot. The number of winning tickets for the public will be Poisson distributed with l 5 n>7,000,000. However, any member of the public who wins will necessarily have to share the jackpot with the syndicate, which is a sure winner. Use this information to calculate the expected amount the public will win. Then do the same calculation when the syndicate does not play. (In this case the jackpot will be smaller, but the public won’t have to share any winnings with the syndicate.) For values of n and R that you can select, is the public better off with or without the syndicate? Would you, as a general member of the public, support a move to outlaw syndicates from hog- ging the tickets?

The second question is whether the syndicate is wise to buy so many tickets. Again assume that the syndicate can spend $7 million and purchase each possible ticket. (Would this be possible in reality?) Also, assume that n members of the general public purchase tickets, and that the carryover

from the previous jackpot is R. The syndicate is thus assured of having a winning ticket, but is it assured of covering its costs? Calculate the expected net benefit (in terms of net present value) to the syndicate, using any reasonable values of n and R, to see whether the syndicate can expect to come out ahead.

Actually, the analysis suggested in the previous para- graph is not complete. There are at least two complications to consider. The first is the effect of taxes. Fortunately for the Australian syndicate, it did not have to pay federal or state taxes on its winnings, but a U.S. syndicate wouldn’t be so lucky. Second, the jackpot from a $20 million jackpot, say, is actually paid in 20 annual $1 million payments. The Lot- tery Commission pays the winner $1 million immediately and then purchases 19 “strips” (bonds with the interest not included) maturing at 1-year intervals with face value of $1 million each. Unfortunately, the lottery prize does not offer the liquidity of the Treasury issues that back up the payments. This lack of liquidity could make the lottery less attractive to the syndicate.

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